# The Schwarz-Christoffel Conformal Mapping for “Polygons” with Infinitely Many Sides

Authors: Gonzalo Riera; Hernán Carrasco; Rubén Preiss
Journal: International Journal of Mathematics and Mathematical Sciences
 (2008)

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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.


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## Body

## 1. Introduction

LetD 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 (1.1)f(z)=α∫0zdx(x−ai)μ1⋯(x−an)μn+β, where a1<⋯<an are real and α,β∈ℂ.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
(1.2)g(z)=α∫1zeλx(x−a1)μ1⋯(x−an)μndxx2+β with λ=∑i=1nμi/ai.In this paper,
we explore the case in which there are infinitely many pointsai 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 toπ/2 and 3π/2. The formula in this case is (1.3)f(z)=∫0ztan⁡(x)dx.The zeros of
the denominatorcos⁡(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(1.4)g(z)=∫0z∏n=0+∞(1+x6⋅4n)dx.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.(1.5)
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.)


## 1.1. Polygons
with an Infinite Number of Sides

This is the case, for example, of an
“infinite stairway” with interior angles alternately equal toπ/2 and 3π/2. The formula in this case is (1.3)f(z)=∫0ztan⁡(x)dx.The zeros of
the denominatorcos⁡(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(1.4)g(z)=∫0z∏n=0+∞(1+x6⋅4n)dx.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.(1.5)
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(2.1)f(z)=∫0ztan⁡(x)dx.The image is given in Figure1,
where the curves shown are the images of horizontal and vertical segments lying
in the half-plane; they can be defined
directly by
(2.2)α(s)=(Re(Numint(tan(x),⁢x=0to0.6+si)),Im(Numint(tan(x),⁢x=0⁢to0.6+si))),0.1≤s≤1.1 for a vertical segment, and by (2.3)β(s)=(Re(Numint(tan(x),x=0to0.09i+s)),Im(Numint(tan(x),x=0to0.09i+s))),0.4≤s≤10 for a horizontal segment.Figure 1
The infinite
staircase.f(z)=∫0ztan⁡xdx.The image of the real axis is the infinite staircase
with steps of equal length(2.4)∫0π/2tan⁡(x)dx=π22.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 Figure2.Figure 2The image of
the positive imaginary axis (labeled (1) in the figure) is given by(2.5)∫0istan⁡(x)dx=i∫0stan⁡(it)dt=−1+i2∫0set−e−tet+e−tdt 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 π2/2 as in Figure 3.Figure 3Consider the triangle of Figure3.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(2.6)f(z)=∫0zsin⁡ν(x)cos⁡μ(x)dx, with 0<ν,μ<1,
generalizes this to zigzag patterns.


### 2.2. The Hairy Half-Plane

We next consider interior anglesπ/2 at ai=(2i+1)π/2(i∈ℤ)and2πatbi=(2i+1)π.The mapping function is therefore(2.7)f(z)=∫0zcos⁡(x/2)cos⁡(x)dx, and the image
is shown in Figure 4.Figure 4
The hairy
half-plane,f(z)=∫0z(cos⁡(x/2)/cos⁡(x))dx.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
angleπ/2 to the tip of the hair is the same on both
sides, that is, that (2.8)∫π/2πcos⁡(x/2)−cos⁡(x)dx=∫π3π/2−cos⁡(x/2)−cos⁡(x), an equality easily obtained by the change of variable x→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 atπ 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 to2π and 0.This means taking(2.9)f(z)=∫0zsin⁡(x)cos⁡(x)dx as in Figure 5.Figure 5
The hairy plane,f(z)=∫0ztan⁡πx⁢dx.


### 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 (2.10)f(z)=∫0zΓ(1/2−x)Γ(−x)dx
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.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 (2.11)1Γ(x)=eγxx∏n=1+∞(1+xn)e−x/n, it follows that (2.12)Γ(1/2−x)Γ(−x)=e−γ/4xx−1/2lim⁡N→+∞∏n=1N(1−x/n1+(1/2−x)/n)1/2en/4.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(2.13)∫mm+1/2Γ(1/2−x)Γ(−x)dx is finite.To obtain a half-stair with the steps moving to the
left one would naturally use(2.14)∫0zΓ(−x)Γ(1/2−x)dx.Finally, to obtain half a hairy plane, we take(2.15)f(z)=∫−1zΓ(1/2−x)Γ(−x)dx as in Figure 7, or
half a hairy half-plane via (2.16)Γ(1/2−x)Γ(−x/2) and so on.Figure 7
Half a hairy
half-plane,f(z)=∫−1z(Γ(1/2−x)/Γ(−x))dx.


## 2.1. The
Infinite Staircase

The first example we consider is the mapping defined
on the upper half-plane by the formula(2.1)f(z)=∫0ztan⁡(x)dx.The image is given in Figure1,
where the curves shown are the images of horizontal and vertical segments lying
in the half-plane; they can be defined
directly by
(2.2)α(s)=(Re(Numint(tan(x),⁢x=0to0.6+si)),Im(Numint(tan(x),⁢x=0⁢to0.6+si))),0.1≤s≤1.1 for a vertical segment, and by (2.3)β(s)=(Re(Numint(tan(x),x=0to0.09i+s)),Im(Numint(tan(x),x=0to0.09i+s))),0.4≤s≤10 for a horizontal segment.Figure 1
The infinite
staircase.f(z)=∫0ztan⁡xdx.The image of the real axis is the infinite staircase
with steps of equal length(2.4)∫0π/2tan⁡(x)dx=π22.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 Figure2.Figure 2The image of
the positive imaginary axis (labeled (1) in the figure) is given by(2.5)∫0istan⁡(x)dx=i∫0stan⁡(it)dt=−1+i2∫0set−e−tet+e−tdt 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 π2/2 as in Figure 3.Figure 3Consider the triangle of Figure3.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(2.6)f(z)=∫0zsin⁡ν(x)cos⁡μ(x)dx, with 0<ν,μ<1,
generalizes this to zigzag patterns.


## 2.2. The Hairy Half-Plane

We next consider interior anglesπ/2 at ai=(2i+1)π/2(i∈ℤ)and2πatbi=(2i+1)π.The mapping function is therefore(2.7)f(z)=∫0zcos⁡(x/2)cos⁡(x)dx, and the image
is shown in Figure 4.Figure 4
The hairy
half-plane,f(z)=∫0z(cos⁡(x/2)/cos⁡(x))dx.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
angleπ/2 to the tip of the hair is the same on both
sides, that is, that (2.8)∫π/2πcos⁡(x/2)−cos⁡(x)dx=∫π3π/2−cos⁡(x/2)−cos⁡(x), an equality easily obtained by the change of variable x→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 atπ 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 to2π and 0.This means taking(2.9)f(z)=∫0zsin⁡(x)cos⁡(x)dx as in Figure 5.Figure 5
The hairy plane,f(z)=∫0ztan⁡πx⁢dx.


## 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 (2.10)f(z)=∫0zΓ(1/2−x)Γ(−x)dx
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.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 (2.11)1Γ(x)=eγxx∏n=1+∞(1+xn)e−x/n, it follows that (2.12)Γ(1/2−x)Γ(−x)=e−γ/4xx−1/2lim⁡N→+∞∏n=1N(1−x/n1+(1/2−x)/n)1/2en/4.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(2.13)∫mm+1/2Γ(1/2−x)Γ(−x)dx is finite.To obtain a half-stair with the steps moving to the
left one would naturally use(2.14)∫0zΓ(−x)Γ(1/2−x)dx.Finally, to obtain half a hairy plane, we take(2.15)f(z)=∫−1zΓ(1/2−x)Γ(−x)dx as in Figure 7, or
half a hairy half-plane via (2.16)Γ(1/2−x)Γ(−x/2) and so on.Figure 7
Half a hairy
half-plane,f(z)=∫−1z(Γ(1/2−x)/Γ(−x))dx.


## 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 Steps1,
2, and 3, and so on.Step 1.
(3.1)Step 2.
(3.2)Step 3.
(3.3)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);(3.4)F(z)=∫1zdx(x−1)2/3(x−exp⁡(2πi/3))2/3(x−exp⁡(4πi/3))2/3=∫1zdx(x3−1)2/3.
This conformal mapping is portrayed in
Figure8.Figure 8
Interior of the Koch
snowflake, Step1.Step 2.
ak=exp⁡(2πik/12)(0≤k≤11);(3.5)αk=13,μk=23,k=0,2,4,68,10αk=43,μk=−13,k=1,3,57,911. The formula for the conformal mapping is (3.6)F(z)=∫1z(x6+1)1/3(x6−1)2/3dx and the image of the unit disc appears in
Figure 9.Figure 9
Interior the Koch snowflake,
Step2.Step 3.
ak=exp⁡(2πik/48)(0≤k<47)(3.7)αk=13,μk=23,k=0,2,6,8,10,14,16,18,22,24,26,30,32,34,38,40,42,46,αk=34,μk=−13allotherk.Although the combinatorics is not automatic, the
formula for the conformal mapping is(3.8)F(z)=∫1z[(x6+1)(x24+1)]1/3[(x3−1)(x3+1)(x12+1)]2/3dx, the image of
the unit disc is given in Figure 10.Figure 10
Interior of Koch's snowflake,
Step3.We may generalize the following.Step 3.1n.
ak=exp⁡(2πik/(3×4n)),0≤k<3×4n with an integrand of the form (3.9)[(x6+1)(x24+1)⋯(x6⋅4n−1+1)]1/3[(x3−1)(x3+1)(x12+1)⋯(x3⋅4n−1+1)]2/3.Theorem 3.1.
The formula for the
conformal map from the interior of the unit disc to the interior of a Koch
snowflake is(3.10)F(z)=∫0z∏n=0+∞(1+x6⋅4n)dx,|z|<1.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
Stepn as (3.11)(1+x6)(1+x24)⋯(1+x6⋅4n−1)(1−x3⋅4n)2/3
(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 Step3:
Define(3.12)a=∫01(1+x24)1/3(1−x24)2/3dx=1.13856,b=∫01x6(1+x24)1/3(1−x24)2/3dx=0.270237, so that (3.13)F(1)=a+b is one vertex
of the polygon. But then (3.14)F(eπi/6)=eπi/6(a−b).
The image of a circular sector from0 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 vertexF(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 (3.15)(a+b)3<a−b,
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 Stepn 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 (3.16)f(z)=∫0z(1+x6)(1+x24)(1+x96)1/3(1−x96)2/3dx.
It is easy to obtain the points(3.17)P*=f(1)=1.35,Q*=f(eπi/24),R*=f(eπy/8),S*=f(eπi/6).
If we compare them to a regular snowflake with
verticesP,⁢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 pointQ* 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 (sinceF′(0)=1).
We check that the image is not all ofℂ.
We claim that(3.18)F(1)=∫01∏n=0+∞(1+x6⋅4n)dx≤34.
Indeed, withy=x6, we have 0≤y≤1 and (3.19)(1+y)(1+y4)(1+y16)⋯=1+y+(y4+y5)+(y16+y17+y20+y21)+⋯≤1+y+2y4+4y16+⋯=1+∑n=0∞2ny4n.
Thus the integral is bounded from above by(3.20)∫011+∑n=0+∞2nx4n⋅6⁢dx=1+∑n=0+∞2n6⋅4n≤1+16∑n=0+∞12n=43.
We also check that the image is not a disc.
Set(3.21)a=∫01∏n=1+∞(1+x6⋅4n)dx,b=∫01x6∏n=1+∞(1+x6⋅4n)dx.
ThenF(1)=a+b and 0<b<a.
But(3.22)F(eπi/6)=eπi/6(a−b) 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 Figure10.
One cannot write, for example,(3.23)(1+x6)1/3(1+x24)1/3=[(1+x6)(1+x24)]1/3 since the
branch we implicitly use by analytic continuation may not coincide with the
fixed branches of the mathematical software.
With(3.24)u(k,x)=log⁡(exp⁡(πik24)−x),v(k,x)=log⁡(x−exp⁡(πik24)),   the numerator
of the integrand is (3.25)Num(x)=exp⁡(1/3(u(1,x)+u(3,x)+u(3,x)+u(4,x)+u(5,x)+u(7,x)+u(9,x)+u(11,x)+v(12,x)+v(13,x)+v(15,x)+v(17,x)+v(19,x)+v(20,x)+v(21,x)+v(23,x)+v(25,x)+v(27,x)+v(28,x)+v(29,x)+v(31,x)+v(33,x)+v(35,x)+v(36,x)+u(37,x)+u(39,x)+u(41,x)+u(43,x)+u(44,x)+u(45,x)+u(47,x))); and the
denominator is (3.26)Den(x)=exp(2/3(u(0,x)+u(2,x)+u(6,x)+u(8,x)+u(10,x)+v(14x)+v(16,x)+v(18,x)+v(22,x)+v(24,x)+v(26,x)+v(30,x)+v(32,x)+v(34,x)+u(38,x)+u(40,x)+u(42,x)+u(46,x))).
At any rate,
the formula of the theorem with terms of up to6⋅46 is sufficiently precise (and simpler to use)
and gives Figure 13.Figure 11Figure 12Figure 13
Exterior of the Koch snowflake, Step1.


### 3.2. The Exterior of the Snowflake

At Step1,
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=13dk=0 so that the formula is (3.27)F(z)=∫1z(x3−1)2/3(dx/x2) whose image is
in Figure 14.Figure 14
Exterior of the Koch snowflake, Step2.In the same way, the formulas for Steps2 and
3 are (3.28)∫1z[(x3−1)(x3+1)]2/3[(x6+1)]1/3dxx2,∫1z[(x3−1)(x3+1)(x12+1)]2/3[(x6+1)(x24+1)]1/3dxx2,   corresponding
to Figures 15 and
16.Figure 15
Exterior of the Koch snowflake, Step3.Figure 16The mapping from the unit disc to the exterior of the
Koch snowflake is given by(3.29)F(z)=∫1z(∏n=0+∞(1+x6⋅4n))−1dxx2.


### 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.(3.30)DefineG(ξ)=∏n=0N(∏k=−∞+∞sin⁡h(∏2h(ξ−an−kaN)))αn.(i)
Guess
initial valuesai,⁢i=1,…,N.(ii)
Locate
the vertices on the triangular contours by(3.31)ZnG=Zn−1G+R∫an−1anG(ξ)dξ 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: (3.32)|annewan−1new|=|anold−an−1old||Zn−Zn−1ZnG−Zn−1G|.This procedure works well in the first few iterations
but, as the authors explain, is computationally prohibitive already form=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
Figure16. For m=1 consider the function (3.33)F(z)=∫0z(1−e2πix/3)(1−e2πi)1/3dx.It maps the upper half plane to a region bounded by
the periodic contour as inR1 (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.Form=2,
the function is defined by (3.34)∫0z(1−e2πix/3)(1−e2πix)1/3(1+e2πix)2/3(1+e4πix)1/3dx and in the limit, for Imz>0,(3.35)∫0z(1−e2πix/3)(1−e2πix)1/3∏n=0+∞(1+e2πi4nx)2/3∏n=0+∞(1+e4πi4nx)1/3dx.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.
(3.36)Step 2.
(3.37)Step 3.
(3.38)We want to define a sequence of Schwarz-Christoffel
mappings from the unit disc to the complements of these trees.The interior angles at Step1 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(3.39)F(z)=∫1z1+x4(1−x4)1/2dxx2 is the required
formula at Step 1.At Step2, we obtain (3.40)F(z)=∫1z1+x4(1−x4)1/21+x8(1+x16)1/2dxx2 with branches
at the 32 roots of unity.Similarly, at Step3(3.41)F(z)=∫1z1+x4(1−x4)1/2(1+x8)(1+x32)[(1+x16)(1+x64)]1/2dxx2.The image is
shown in Figure17.Figure 17
Exterior of a tree, Step3.In the limit, we obtain the formula(3.42)F(z)=∫1z1+x4(1−x4)1/2∏n=1+∞(1+x2⋅4n)[∏n=1+∞(1+x4n+1)]1/2dxx2 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.


## 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 Steps1,
2, and 3, and so on.Step 1.
(3.1)Step 2.
(3.2)Step 3.
(3.3)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);(3.4)F(z)=∫1zdx(x−1)2/3(x−exp⁡(2πi/3))2/3(x−exp⁡(4πi/3))2/3=∫1zdx(x3−1)2/3.
This conformal mapping is portrayed in
Figure8.Figure 8
Interior of the Koch
snowflake, Step1.Step 2.
ak=exp⁡(2πik/12)(0≤k≤11);(3.5)αk=13,μk=23,k=0,2,4,68,10αk=43,μk=−13,k=1,3,57,911. The formula for the conformal mapping is (3.6)F(z)=∫1z(x6+1)1/3(x6−1)2/3dx and the image of the unit disc appears in
Figure 9.Figure 9
Interior the Koch snowflake,
Step2.Step 3.
ak=exp⁡(2πik/48)(0≤k<47)(3.7)αk=13,μk=23,k=0,2,6,8,10,14,16,18,22,24,26,30,32,34,38,40,42,46,αk=34,μk=−13allotherk.Although the combinatorics is not automatic, the
formula for the conformal mapping is(3.8)F(z)=∫1z[(x6+1)(x24+1)]1/3[(x3−1)(x3+1)(x12+1)]2/3dx, the image of
the unit disc is given in Figure 10.Figure 10
Interior of Koch's snowflake,
Step3.We may generalize the following.Step 3.1n.
ak=exp⁡(2πik/(3×4n)),0≤k<3×4n with an integrand of the form (3.9)[(x6+1)(x24+1)⋯(x6⋅4n−1+1)]1/3[(x3−1)(x3+1)(x12+1)⋯(x3⋅4n−1+1)]2/3.Theorem 3.1.
The formula for the
conformal map from the interior of the unit disc to the interior of a Koch
snowflake is(3.10)F(z)=∫0z∏n=0+∞(1+x6⋅4n)dx,|z|<1.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
Stepn as (3.11)(1+x6)(1+x24)⋯(1+x6⋅4n−1)(1−x3⋅4n)2/3
(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 Step3:
Define(3.12)a=∫01(1+x24)1/3(1−x24)2/3dx=1.13856,b=∫01x6(1+x24)1/3(1−x24)2/3dx=0.270237, so that (3.13)F(1)=a+b is one vertex
of the polygon. But then (3.14)F(eπi/6)=eπi/6(a−b).
The image of a circular sector from0 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 vertexF(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 (3.15)(a+b)3<a−b,
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 Stepn 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 (3.16)f(z)=∫0z(1+x6)(1+x24)(1+x96)1/3(1−x96)2/3dx.
It is easy to obtain the points(3.17)P*=f(1)=1.35,Q*=f(eπi/24),R*=f(eπy/8),S*=f(eπi/6).
If we compare them to a regular snowflake with
verticesP,⁢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 pointQ* 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 (sinceF′(0)=1).
We check that the image is not all ofℂ.
We claim that(3.18)F(1)=∫01∏n=0+∞(1+x6⋅4n)dx≤34.
Indeed, withy=x6, we have 0≤y≤1 and (3.19)(1+y)(1+y4)(1+y16)⋯=1+y+(y4+y5)+(y16+y17+y20+y21)+⋯≤1+y+2y4+4y16+⋯=1+∑n=0∞2ny4n.
Thus the integral is bounded from above by(3.20)∫011+∑n=0+∞2nx4n⋅6⁢dx=1+∑n=0+∞2n6⋅4n≤1+16∑n=0+∞12n=43.
We also check that the image is not a disc.
Set(3.21)a=∫01∏n=1+∞(1+x6⋅4n)dx,b=∫01x6∏n=1+∞(1+x6⋅4n)dx.
ThenF(1)=a+b and 0<b<a.
But(3.22)F(eπi/6)=eπi/6(a−b) 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 Figure10.
One cannot write, for example,(3.23)(1+x6)1/3(1+x24)1/3=[(1+x6)(1+x24)]1/3 since the
branch we implicitly use by analytic continuation may not coincide with the
fixed branches of the mathematical software.
With(3.24)u(k,x)=log⁡(exp⁡(πik24)−x),v(k,x)=log⁡(x−exp⁡(πik24)),   the numerator
of the integrand is (3.25)Num(x)=exp⁡(1/3(u(1,x)+u(3,x)+u(3,x)+u(4,x)+u(5,x)+u(7,x)+u(9,x)+u(11,x)+v(12,x)+v(13,x)+v(15,x)+v(17,x)+v(19,x)+v(20,x)+v(21,x)+v(23,x)+v(25,x)+v(27,x)+v(28,x)+v(29,x)+v(31,x)+v(33,x)+v(35,x)+v(36,x)+u(37,x)+u(39,x)+u(41,x)+u(43,x)+u(44,x)+u(45,x)+u(47,x))); and the
denominator is (3.26)Den(x)=exp(2/3(u(0,x)+u(2,x)+u(6,x)+u(8,x)+u(10,x)+v(14x)+v(16,x)+v(18,x)+v(22,x)+v(24,x)+v(26,x)+v(30,x)+v(32,x)+v(34,x)+u(38,x)+u(40,x)+u(42,x)+u(46,x))).
At any rate,
the formula of the theorem with terms of up to6⋅46 is sufficiently precise (and simpler to use)
and gives Figure 13.Figure 11Figure 12Figure 13
Exterior of the Koch snowflake, Step1.


## 3.2. The Exterior of the Snowflake

At Step1,
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=13dk=0 so that the formula is (3.27)F(z)=∫1z(x3−1)2/3(dx/x2) whose image is
in Figure 14.Figure 14
Exterior of the Koch snowflake, Step2.In the same way, the formulas for Steps2 and
3 are (3.28)∫1z[(x3−1)(x3+1)]2/3[(x6+1)]1/3dxx2,∫1z[(x3−1)(x3+1)(x12+1)]2/3[(x6+1)(x24+1)]1/3dxx2,   corresponding
to Figures 15 and
16.Figure 15
Exterior of the Koch snowflake, Step3.Figure 16The mapping from the unit disc to the exterior of the
Koch snowflake is given by(3.29)F(z)=∫1z(∏n=0+∞(1+x6⋅4n))−1dxx2.


## 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.(3.30)DefineG(ξ)=∏n=0N(∏k=−∞+∞sin⁡h(∏2h(ξ−an−kaN)))αn.(i)
Guess
initial valuesai,⁢i=1,…,N.(ii)
Locate
the vertices on the triangular contours by(3.31)ZnG=Zn−1G+R∫an−1anG(ξ)dξ 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: (3.32)|annewan−1new|=|anold−an−1old||Zn−Zn−1ZnG−Zn−1G|.This procedure works well in the first few iterations
but, as the authors explain, is computationally prohibitive already form=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
Figure16. For m=1 consider the function (3.33)F(z)=∫0z(1−e2πix/3)(1−e2πi)1/3dx.It maps the upper half plane to a region bounded by
the periodic contour as inR1 (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.Form=2,
the function is defined by (3.34)∫0z(1−e2πix/3)(1−e2πix)1/3(1+e2πix)2/3(1+e4πix)1/3dx and in the limit, for Imz>0,(3.35)∫0z(1−e2πix/3)(1−e2πix)1/3∏n=0+∞(1+e2πi4nx)2/3∏n=0+∞(1+e4πi4nx)1/3dx.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.
(3.36)Step 2.
(3.37)Step 3.
(3.38)We want to define a sequence of Schwarz-Christoffel
mappings from the unit disc to the complements of these trees.The interior angles at Step1 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(3.39)F(z)=∫1z1+x4(1−x4)1/2dxx2 is the required
formula at Step 1.At Step2, we obtain (3.40)F(z)=∫1z1+x4(1−x4)1/21+x8(1+x16)1/2dxx2 with branches
at the 32 roots of unity.Similarly, at Step3(3.41)F(z)=∫1z1+x4(1−x4)1/2(1+x8)(1+x32)[(1+x16)(1+x64)]1/2dxx2.The image is
shown in Figure17.Figure 17
Exterior of a tree, Step3.In the limit, we obtain the formula(3.42)F(z)=∫1z1+x4(1−x4)1/2∏n=1+∞(1+x2⋅4n)[∏n=1+∞(1+x4n+1)]1/2dxx2 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.



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*Source: 350326-2008-08-31.xml*