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Gonzalo Riera 1 and Hernán Carrasco 1 and Rubén Preiss 1
Recommended by Vladimir Mityushev
1, Departamento de Matemáticas, Pontificia Universidad Católica de Chile, Avenue Vicuña Makenna 4860, 7820436 Macul, Santiago, Chile
Received 22 January 2008; Revised 15 May 2008; Accepted 1 July 2008
1. Introduction
Let D be a polygon in the complex plane with n vertices and interior angles παi ,... 0≤αi ≤2, 1≤i≤n ; the exterior angles are given by πμi , where αi +μi =1 . The Schwarz-Christoffel mapping of the upper-half plane onto D is effected by a function of the form [figure omitted; refer to PDF] where a1 <...<an are real and α,β∈... .
Although there is ample literature on the subject (see, e.g., [1-3], as well as the complete overview contained in Driscoll and Threfethen [4]), a few remarks would not be out of place here. First of all, even though (1.1) gives the appearance of being an explicit formula, there is actually no known relation between the values of the ai and the lengths of the sides of D . For this reason, when considering an infinite-sided polygon, one needs to justify passage to a corresponding infinite product, which is what we do in what follows. Also, formula (1.1) is necessary but not sufficient in the sense that for some values of the ai 's, the mapping need not to be one-to-one (see [5]). To establish the necessity, one needs to consider the behavior of f(z) at ∞ , a possibility also clearly excluded in the case of infinitely many ai .
Before proceeding, we review some variations of (1.1). One of the points ai may be located at ∞ , the formula remaining the same, and one of the vertices of the polygon may be at ∞ , where we use the relation μ1 +...+μn =2 . The formula for mapping the unit disc conformally onto D is essentially the same, but with |ai |=1, 1≤i≤n . The formula for the mapping onto the exterior of the polygon is [figure omitted; refer to PDF] with λ=∑i=1n μi / ai .
In this paper, we explore the case in which there are infinitely many points ai on the real line (or the unit circle) and obtain results for two kinds of "polygons."
1.1. Polygons with an Infinite Number of Sides
This is the case, for example, of an "infinite stairway" with interior angles alternately equal to π/2 and 3π/2 . The formula in this case is [figure omitted; refer to PDF]
The zeros of the denominator cos...(x) correspond to ai =(2i+1)π/2, μi =1/2 and those of the numerator sin...(x) to ai =iπ,... μi =-1/2 .
1.2. Fractals
This is the case, for example, of the interior of the Koch snowflake, for which the formula on the interior of the unit disc turns out to be [figure omitted; refer to PDF]
A similar generalization of (1.2) gives the conformal mapping onto the exterior of the Koch snowflake.
To reproduce the figures in the text, a program such as Derive, Maple, or Mathematica is required; we explain in each case how to obtain these figures.
To our knowledge, formulas of this kind have not previously been considered in the literature, and the initial nature of the present investigation precludes exhaustive results. It is our hope that others will be motivated to seek general theorems along these lines, or at any rate to find explicit mappings of the upper-half-plane onto other infinitely sided polygons. We mention here the open problem of a half-strip slit along infinite sequences of equal-length line segments emanating from the infinite sides. [figure omitted; refer to PDF] The mapping of the half-plane onto this domain is of importance in the study of differentials on hyperelliptic surfaces of infinite genus as considered in the pioneering work by Myrberg, see [5]. (The formula for the case of the analogously slitted full strip is obtained below.)
2. Polygons with an Infinite Number of Sides
2.1. The Infinite Staircase
The first example we consider is the mapping defined on the upper half-plane by the formula [figure omitted; refer to PDF]
The image is given in Figure 1, where the curves shown are the images of horizontal and vertical segments lying in the half-plane; they can be defined directly by [figure omitted; refer to PDF] for a vertical segment, and by [figure omitted; refer to PDF] for a horizontal segment.
Figure 1: The infinite staircase. f(z)=∫0z tan... x dx.
[figure omitted; refer to PDF]
The image of the real axis is the infinite staircase with steps of equal length [figure omitted; refer to PDF]
In order to prove that this formula actually defines a one-to-one mapping, we proceed as follows.
Consider the triangle (i.e., half-strip) of Figure 2.
Figure 2: [figure omitted; refer to PDF]
The image of the positive imaginary axis (labeled (1) in the figure) is given by [figure omitted; refer to PDF] for 0≤s≤+∞ , so that it is the half-line making an angle of 3π/4 with the positively oriented real axis. The image of (2) is the real segment [0,π2/2] , and that image of (3) is a half-line parallel to the first one and beginning at π 2/2 as in Figure 3.
Figure 3: [figure omitted; refer to PDF]
Consider the triangle of Figure 3.
The general theory of conformal mappings applies to these two triangles, so that from the bijectivity on the boundaries we can conclude bijectivity in the interior. With the Schwarz reflection principle as applied to the vertical segments (1), (3), and their reflections, the mapping can now be extended to the entire upper half-plane.
The function [figure omitted; refer to PDF] with 0<ν, μ<1 , generalizes this to zigzag patterns.
2.2. The Hairy Half-Plane
We next consider interior angles π/2 at ai =(2i+1)π/2 (i∈...) and 2π at bi =(2i+1)π.
The mapping function is therefore [figure omitted; refer to PDF] and the image is shown in Figure 4.
Figure 4: The hairy half-plane, f(z)=∫0z (cos...(x/2)/cos...(x)) dx.
[figure omitted; refer to PDF]
To establish that the given mapping is indeed one-to-one, it is first necessary to see that the length of the segment from an angle π/2 to the tip of the hair is the same on both sides, that is, that [figure omitted; refer to PDF] an equality easily obtained by the change of variable x[arrow right]2π-x .
The proof can now be completed in the same way as for the infinite staircase by first considering the image of a triangle with vertical sides at π and 3π , and then extending the mapping by the reflection principle.
In a similar fashion, we obtain a formula for the hairy plane, which can be regarded as a polygon with infinitely many sides and interior angles alternately equal to 2π and 0 .
This means taking [figure omitted; refer to PDF] as in Figure 5.
Figure 5: The hairy plane, f(z)=∫0z tan...πx... dx.
[figure omitted; refer to PDF]
2.3. The Infinite Half-Staircase
In order to consider only one half of the stair, we have to take analytic functions in (1.1) having half as many zeros as before, which leads to the Γ function. Indeed, the formula [figure omitted; refer to PDF] defines such a mapping (see Figure 6).
Figure 6: Half a stair or the stairway to the abyss, f(z)=∫0z (Γ(1/2-x)/Γ(-x)) dx.
[figure omitted; refer to PDF]
However, to prove that the mapping given in (2.10) is one-to-one, we cannot use the arguments of examples 1 and 2 since there is no symmetry with respect to a line. We proceed as follows. Since [figure omitted; refer to PDF] it follows that [figure omitted; refer to PDF]
The classical Schwarz-Christoffel formula can be applied to each finite product on the right-hand side, which gives a staircase with a finite number of steps. These conformal mappings converge uniformly to the required mapping since for each positive integer, the integral [figure omitted; refer to PDF] is finite.
To obtain a half-stair with the steps moving to the left one would naturally use [figure omitted; refer to PDF]
Finally, to obtain half a hairy plane, we take [figure omitted; refer to PDF] as in Figure 7, or half a hairy half-plane via [figure omitted; refer to PDF] and so on.
Figure 7: Half a hairy half-plane, f(z)=∫-1z (Γ(1/2-x)/Γ(-x)) dx.
[figure omitted; refer to PDF]
3. Fractals
3.1. The Koch Snowflake
Driscoll and Trefethen ([4, Chapter 4.5]) remark that the conformal mapping into the interior or exterior of complicated fractals, such as the Koch snowflake, ought to be treatable in a manner related to the multiple method. In the next theorem, we give an explicit geometric formula which does this.
Numerical schemes have been studied before using iteration procedures, see [6].
The Koch snowflake is obtained as the limit of a sequence of polygons defined as in Steps 1, 2, and 3, and so on.
Step 1.
[figure omitted; refer to PDF]
Step 2.
[figure omitted; refer to PDF]
Step 3.
[figure omitted; refer to PDF]
At each step, we consider the Schwarz-Christoffel mapping of the unit disc into the interior of the polygon; we use the notation of the introduction throughout.
Step 1.
αk =1/3, μk =2/3, ak =exp(2πik/3) (k=1,2,3); [figure omitted; refer to PDF]
This conformal mapping is portrayed in Figure 8.
Figure 8: Interior of the Koch snowflake, Step 1.
[figure omitted; refer to PDF]
Step 2.
ak =exp...(2πik/12) (0≤k≤11); [figure omitted; refer to PDF] The formula for the conformal mapping is [figure omitted; refer to PDF] and the image of the unit disc appears in Figure 9.
Figure 9: Interior the Koch snowflake, Step 2.
[figure omitted; refer to PDF]
Step 3.
ak =exp...(2πik/48) (0≤k<47) [figure omitted; refer to PDF]
Although the combinatorics is not automatic, the formula for the conformal mapping is [figure omitted; refer to PDF] the image of the unit disc is given in Figure 10.
Figure 10: Interior of Koch's snowflake, Step 3.
[figure omitted; refer to PDF]
We may generalize the following.
Step 3.1 n .
ak =exp...(2πik/(3×4n )), 0≤k<3×4n with an integrand of the form [figure omitted; refer to PDF]
Theorem 3.1.
The formula for the conformal map from the interior of the unit disc to the interior of a Koch snowflake is [figure omitted; refer to PDF]
Proof.
The wording "a Koch snowflake" means that it is the limit of polygons with the same angles as the polygons in the Koch pattern, but with unequal sides.
First, we rewrite the integrand for Step n as [figure omitted; refer to PDF] (modulo a cube root of 1), so that for |x|<1 the integrand converges to the one indicated in the statement of the theorem.
That the lengths are not equal to the lengths in true Koch snowflake can already be seen at Step 3:
Define [figure omitted; refer to PDF] so that [figure omitted; refer to PDF] is one vertex of the polygon. But then [figure omitted; refer to PDF]
The image of a circular sector from 0 to 1 to exp...(πi/6) is therefore a polygon of the form of Figure 11.
If the lengths were equal to those of the regular snowflake then the vertex F(exp...(πi/6)) would be the vertex of the isosceles triangle with base a+b and base angle π/6 .
The side would equal (a+b)/3 . But [figure omitted; refer to PDF] showing that the second length in the small triangle is necessarily shorter than the first one, leading to a location of F(exp...(πi/6)) slightly "higher" than it should be.
Observe also that the location of this last vertex implies (the angles being as required) that the map is one-to-one on the boundary and therefore one-to-one throughout.
In general, the image in Step n differs somewhat, but not by too much, from a regular snowflake with all sides equal.
If we analyze a sufficiently general case as in Step3, say that we have the mapping [figure omitted; refer to PDF]
It is easy to obtain the points [figure omitted; refer to PDF]
If we compare them to a regular snowflake with vertices P,... Q,... R,... S , we obtain points at a distance smaller than 0.1 .
What is to be noticed is that for the image not be one-to-one on the boundary, the point Q* should differ sufficiently from Q , to be below PO , for example; the position of P* ,... Q* ,... R* ,... S* thus forces the actual polygon to be free of self-intersections, the angles being π/3 or 2π/3 . We draw one such possible polygon in Figure 12.
The sequence of conformal mappings therefore converges to a nonconstant conformal mapping (since F[variant prime] (0)=1 ).
We check that the image is not all of ... .
We claim that [figure omitted; refer to PDF]
Indeed, with y=x6 , we have 0≤y≤1 and [figure omitted; refer to PDF]
Thus the integral is bounded from above by [figure omitted; refer to PDF]
We also check that the image is not a disc.
Set [figure omitted; refer to PDF]
Then F(1)=a+b and 0<b<a .
But [figure omitted; refer to PDF] has modulus smaller than F(1) .
Finally, we have to explain a subtle point that arises in producing a drawing such as the one in Figure 10.
One cannot write, for example, [figure omitted; refer to PDF] since the branch we implicitly use by analytic continuation may not coincide with the fixed branches of the mathematical software.
With [figure omitted; refer to PDF] the numerator of the integrand is [figure omitted; refer to PDF] and the denominator is [figure omitted; refer to PDF]
At any rate, the formula of the theorem with terms of up to 6[sm middot]46 is sufficiently precise (and simpler to use) and gives Figure 13.
Figure 11: [figure omitted; refer to PDF]
Figure 12: [figure omitted; refer to PDF]
Figure 13: Exterior of the Koch snowflake, Step 1.
[figure omitted; refer to PDF]
3.2. The Exterior of the Snowflake
At Step 1, we use the formula of the introduction with ak =exp...(2πi/3×k), μk =2/3, 1≤k≤3 .
In this case, λ=2/3 ∑k=13 dk =0 so that the formula is [figure omitted; refer to PDF] whose image is in Figure 14.
Figure 14: Exterior of the Koch snowflake, Step 2.
[figure omitted; refer to PDF]
In the same way, the formulas for Steps 2 and 3 are [figure omitted; refer to PDF] corresponding to Figures 15 and 16.
Figure 15: Exterior of the Koch snowflake, Step 3.
[figure omitted; refer to PDF]
Figure 16: [figure omitted; refer to PDF]
The mapping from the unit disc to the exterior of the Koch snowflake is given by [figure omitted; refer to PDF]
3.3. Periodic Koch Fractals
In a paper on diffusive transport by Brady and Pozrikidis (cf. [6]) the authors consider conformal mappings of a rectangle onto regions Rm ,... m=1,2,... as in Figure 16.
The mappings can then be extended by reflections on the vertical sides to mappings from an infinite strip to regions bounded below by the given contours. In the limit, a fractal region is obtained.
The mapping is given by an iteration procedure which we summarize as follows. [figure omitted; refer to PDF]
(i) Guess initial values ai ,... i=1,...,N .
(ii) Locate the vertices on the triangular contours by [figure omitted; refer to PDF] and compare them to the known exact values of the vertices an . If they are equal the calculation stops; otherwise compute improved values for an using the corrector formula: [figure omitted; refer to PDF]
This procedure works well in the first few iterations but, as the authors explain, is computationally prohibitive already for m=5 (cf. [6] for details).
Using the Schwarz-Christoffel formula in a similar way as in number 2 above, we may define a conformal mapping to a region bounded below by a periodic fractal of Koch's type obtained from Figure 16. For m=1 consider the function [figure omitted; refer to PDF]
It maps the upper half plane to a region bounded by the periodic contour as in R1 (extended by reflection on the vertical sides).
The actual lengths of the sides in the image are {1,2,1} so that they do not coincide with preassigned value 1 for all sides, but the image has the same geometrical shape.
For m=2 , the function is defined by [figure omitted; refer to PDF] and in the limit, for Im z>0 , [figure omitted; refer to PDF]
The infinite products converge uniformly in compact sets on the upper half plane and the integral defines a conformal map onto a region bounded below by an infinite periodic fractal of Koch type. As usual with these maps, we cannot so far control precisely the lengths of the sides except numerically for the first few steps.
3.4. A Tree
We define a sequence of "trees" as follows.
Step 1.
[figure omitted; refer to PDF]
Step 2.
[figure omitted; refer to PDF]
Step 3.
[figure omitted; refer to PDF]
We want to define a sequence of Schwarz-Christoffel mappings from the unit disc to the complements of these trees.
The interior angles at Step 1 give α1 =α3 =α5 =α7 =0 with μ1 =μ3 =μ5 =μ7 =1 at the points and α2 =α4 =α6 =α8 =3/2, μ2 =μ4 =μ6 =μ8 =-1/2 at the corners.
Therefore [figure omitted; refer to PDF] is the required formula at Step 1.
At Step 2, we obtain [figure omitted; refer to PDF] with branches at the 32 roots of unity.
Similarly, at Step 3 [figure omitted; refer to PDF]
The image is shown in Figure 17.
Figure 17: Exterior of a tree, Step 3.
[figure omitted; refer to PDF]
In the limit, we obtain the formula
[figure omitted; refer to PDF] for the conformal mapping of the unit disc onto the exterior of this fractal.
There are of course many more trees to be considered but the combinatorics arising from angles and points is not straightforward, so that it may be hard to obtain a general expression for the conformal mapping. We leave these as questions for further study.
5. Conclusion
The classical Schwarz-Christoffel formula maps the upper half plane to a polygon with prescribed angles, but with sides of unknown lengths. Notwithstanding this, we have shown that the formula can be extended to a variety of shapes with an infinite number of vertices, namely, stairs, slitted planes, fractals of Koch's type and trees. It would be very interesting in the future to find exactly the branch points necessary in Koch snowflake as well as in specific slitted regions.
Acknowledgment
This research was partially suported by Fondecyt 1050904.
[1] D. Bonciani, F. Vlacci, "Some remarks on Schwarz-Christoffel transformations from the unit disk to a regular polygon and their numerical computation," Complex Variables and Elliptic Equations , vol. 49, no. 4, pp. 271-284, 2004.
[2] P. Henrici Applied and Computational Complex Analysis. Vol. 1: Power Series, Integration, Conformal Mapping, Location of Zeros , of Pure and Applied Mathematics, pp. xv+682, John Wiley & Sons, New York, NY, USA, 1974.
[3] E. Johnston, "A "counter example" for the Schwarz-Christoffel transform," The American Mathematical Monthly , vol. 90, no. 10, pp. 701-703, 1983.
[4] T. A. Driscoll, L. N. Trefethen Schwarz-Christoffel Mapping , vol. 8, of Cambridge Monographs on Applied and Computational Mathematics, pp. xvi+132, Cambridge University Press, Cambridge, UK, 2002.
[5] P. J. Myrberg, "Über analytische Funktionen auf transzendenten zweiblättrigen Riemannschen Flächen mit reellen Verzweigungspunkten," Acta Mathematica , vol. 76, no. 3-4, pp. 185-224, 1945.
[6] M. Brady, C. Pozrikidis, "Diffusive transport across irregular and fractal walls," Proceedings of the Royal Society of London. Series A , vol. 442, no. 1916, pp. 571-583, 1993.
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Abstract
The classical Schwarz-Christoffel formula gives conformal mappings of the upper half-plane onto domains whose boundaries consist of a finite number of line segments. In this paper, we explore extensions to boundary curves which in one sense or another are made up of infinitely many line segments, with specific attention to the "infinite staircase" and to the Koch snowflake, for both of which we develop explicit formulas for the mapping function and explain how one can use standard mathematical software to generate corresponding graphics. We also discuss a number of open questions suggested by these considerations, some of which are related to differentials on hyperelliptic surfaces of infinite genus.
You have requested "on-the-fly" machine translation of selected content from our databases. This functionality is provided solely for your convenience and is in no way intended to replace human translation. Show full disclaimer
Neither ProQuest nor its licensors make any representations or warranties with respect to the translations. The translations are automatically generated "AS IS" and "AS AVAILABLE" and are not retained in our systems. PROQUEST AND ITS LICENSORS SPECIFICALLY DISCLAIM ANY AND ALL EXPRESS OR IMPLIED WARRANTIES, INCLUDING WITHOUT LIMITATION, ANY WARRANTIES FOR AVAILABILITY, ACCURACY, TIMELINESS, COMPLETENESS, NON-INFRINGMENT, MERCHANTABILITY OR FITNESS FOR A PARTICULAR PURPOSE. Your use of the translations is subject to all use restrictions contained in your Electronic Products License Agreement and by using the translation functionality you agree to forgo any and all claims against ProQuest or its licensors for your use of the translation functionality and any output derived there from. Hide full disclaimer