Fractal Transformations on Harmonic Functions

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    Fractal transformations of harmonic functions

    Michael F. Barnsley, Uta FreibergDepartment of Mathematics, Australian National University, Canberra, ACT, Australia

    28 December 2006

    ABSTRACT

    The theory of fractal homeomorphisms is applied to transform a Sierpinski triangle into what we call a Kigamitriangle. The latter is such that the corresponding harmonic functions and the corresponding Laplacian 4 take arelatively simple form. This provides an alternative approach to recent results of Teplyaev. Using a second fractalhomeomorphism we prove that the outer boundary of the Kigami triangle possesses a continuous first derivativeat every point. This paper shows that IFS theory and the chaos game algorithm provide important tools foranalysis on fractals.

    1 Introduction

    We begin this paper with a brief review of the recently developed theory of fractal tops and fractal transforma-

    tions. We emphasize that fractal transformations may be computed readily by means of the chaos game algorithm.Then we develop a beautiful application: we show how the theory may be applied to transform the Sierpinskitriangle so that the corresponding harmonic functions and the corresponding Laplacian 4take a relatively simpleform. This provides an alternative approach to recent results of Teplyaev31 28 29 30 . In particular, Kigami1617 appears to have written the first papers regarding representation of Sierpinski triangle in harmonic coordinatesand Meyers23 may have been to present the geometrical interpretation of the harmonic representation. Here weprove, by use the fractal homeomorphism theorem a second time and a third time, that the basic curves, fromwhich the Kigami triangle may be constructed, are continuously differentiable.

    Other relevant references of which we are aware are Berger9 , Kusuoka18 19 20 , and also Strichartz26 ,top of page 194, where there is mention of three-by-three row-stochastic transformations in equations (5) andtheir relation to the corresponding two-by-two matrices in equations (8), with mention of the eigenvectors andeigenvalues of the latter. See also Barnsley4 5 concerning the impedence functions and spectra of renormalizable

    electro-mechanical systems and their relation to Julia sets.

    2 Hyperbolic IFS and Birkhoffs ergodic theorem

    Definition 1. Let (X, dX) be a complete metric space. Let {f1, f2,...,fN} be a finite sequence of strictly

    1

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    contractive transformations, fn: X X, forn = 1, 2,...,N. Then

    F :={X; f1, f2,...,fN}

    is called a hyperbolic iterated function system or hyperbolic IFS.

    A transformation fn: X X is strictly contractive iffthere exists a number ln [0, 1)such that d(fn(x), fn(y)) lnd(x, y)for allx, y X.

    Let denote the set of all infinite sequences{k}k=1of symbols belonging to the alphabet {1,...,N}. We write = 123... to denote a typical element of, and we write k to denote the k th element of . Then(, d)is a compact metric space, where the metric dis defined byd(, ) = 0when = and d(, ) = 2k

    when k is the least index for which k 6=k. We call the code space associated with the IFS F.

    Let and x X. Then, using the contractivity ofF, it is straightfoward to prove that

    F() := limk

    f1 f2 ...fk(x) (1)

    exists, uniformly for x in any fixed compact subset of X, and depends continuously on . See for example1 ,

    Theorem 3. LetAF={F() : }.

    Then AF X is called the attractor ofF. The continuous function

    F : AF

    is called the address function ofF. We call 1F ({x}) := { : F() = x} the set of addresses of the pointx AF.

    ClearlyAFis compact, nonempty, and has the property

    AF=f1(AF) f2(AF) ... fN(AF).

    Indeed, if we define H(X)to be the set of nonempty compact subsets ofX, and we define F : H(X) H(X)byF(S) = f1(S) f2(S) ... fN(S), (2)

    for all S H(X), then AFcan be characterized as the unique fixed point ofF, see Hutchinson14 , section 3.2,and Williams32 .

    IFSs may be used to represent diverse subsets ofR2. An IFS whose attractor is the Sierpinski triangle withvertices atA = (0, 0),B = (1, 0), and C= (0.5,

    p0.75))is

    F={R2; f1, f2, f3}

    wheref1(x) = 12 (x +A), f2(x) = 1

    2(x +B),and f3(x) = 12 (x +C) with x= (x, y). We may representf1 using

    matrix notation for affi

    ne transformations, namely,1

    2 0

    0 12

    xy

    +

    AxAy

    ,

    and similarly for f2, f3. We will write ABCto denote the triangle whose vertices are A,B, and C. Notice thatABC is an equilateral triangle.

    Definition 2. An IFS with probabilities is a hyperbolic IFS {X; f1, f2,...,fN} together with a set of realnumberspn> 0 forn = 1, 2,...,N withp1+p2+ ... +pN= 1.

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    It is a basic result of IFS theory that given any IFS Fwith probabilities there exists a unique Borel measureof norm one, namely a probility measure, such that

    (B) =NPn=1

    pn(f1i (B))

    for all Borel subsets B ofX

    . The support of this measure is the attractor AFof the IFS. This measure may bereferred to as the measure-attractorof the IFS. The standard measure on the Sierpinski triangle corresponds to

    p1 = p2= p3=1

    3.

    This is a normalized ln 3/ ln 2- dimensional Hausdorffmeasure.

    Both the measure-attractor and the attractor of an IFS can be described in terms of thechaos game algorithm.We define a random orbit ofx0 Xto be {xk}k=0 where

    xk+1= fk(xk),

    k {1, 2,...,N} with k = n with probability pn independently of all other choices. (That is, the sequence123... is i.i.d.) Then, with probability one, the sequence of measures

    k := 1

    k

    kPj=1

    xj

    converges weakly to and, also with probability one, the intersection of the nested (decreasing) sequence ofcompact sets

    Ak= {xj}j=k

    equalsAF. The bar denotes closure.

    The proof of these results relies on Birkhoffs ergodic theorem, see for example Vrscay12 . The scholarlyhistory of the chaos game is discussed by Kaijer15 and Stenflo25 , and appears to begin in 1935 with the work ofOnicescu and Mihok24 . Mandelbrot22 , pp.196-199, used a version of it to help compute pictures of certain Juliasets; it was introduced to IFS theory and developed by one of the authors and his coworkers, see for exampleBarnsley and Demko1 , Barnsley6 , Berger10 , and Elton11 , where the relevant theorems and much discussioncan be found.

    3 Homeomorphisms between fractals

    We order the elements ofaccording to

    < iffk > k

    where k is the least index for which k 6= k. This is a linear ordering, sometimes called the lexicographic

    ordering.Notice that all elements of are less than or equal to 1 = 11111... and greater than or equal to N =

    NNNNN..... Also, any pair of distinct elements of is such that one member of the pair is strictly greater thanthe other. In particular, the set of addresses of a pointx AFis both closed and bounded above by 1. It followsthat1F ({x})possesses a unique largest element. We denote this element by F(x).

    Definition 1. LetFbe a hyperbolic IFS with attractorAF and address functionF : AF. LetF(x) = max{ :F() = x} for allx AF.

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    Then

    F :={F(x) : x AF}is called the tops code space and

    F :AF Fis called the tops function, for the IFSF.

    We remark that Fis a shift invariant subspace , see Barnsley7 . Consequently the rich theory of symbolicdynamics, see Lind21 , can be brought to bear on the analysis of tops code spaces and can provide much informationabout the underlying fractal structures.

    But we will need the following. Let G denote a hyperbolic IFS that also consists of N functions. ThenG F :AF AG is a mapping from the attractor ofFinto the attractor ofG . We refer to G Fas a fractaltransformation.

    Definition 2. The address structure ofF is defined to be the set of sets

    CF={1F ({x}) F :x AF}.

    The address structure of an IFS is a certain partition ofF. Let CG denote the address structure ofG. Letus write CF CG to mean that for each S CFthere is T CG such that S T. Notice that ifCF = CG thenF= G. Some examples of address structures are given by Barnsley8 .

    Theorem 3. Let F and G be two hyperbolic IFSs such that CF CG. Then the fractal transformationG F :AF AG is continuous. IfCF=CG thenG F is a homeomorphism.

    The proof is given by Barnsley7 8 .

    Given the two hyperbolic IFSs F and G, such that CF = CG , we note that the attractor of the IFS F Gdefined by

    F G = {X X; wn = (fn, gn), n= 1, 2,...,N}.

    contains the graphGof the homeomorphismG F. Loosely speaking, the "top" of the attractor ofF GequalsG. Precisely, ifAF is totally disconnected or finitely ramified then AFG =G; in general it is straightfoward toextractG from AFG . But the point we want to emphasize here is that G can be computed by means of thechaos game algorithm applied to the IFS F G, see Barnsley7 .

    4 Energy and harmonic functions on a sierpinki triangle

    LetF={R2; f1, f2, f3}denote the IFS associated with the Sierpinski triangle AFwith vertices at A, B,andC. LetV0 denote the set of vertices of the triangle ABC. Namely,V0= {A , B, C}.

    Let0

    denote the set offi

    nite sequences of symbols from the alphabet {1, 2,...,N}, including the empty string.That is,

    0 =

    Sk=1

    {1, 2, 3}k {}.

    Let 0. We write ||to denote the length of . Then is the empty string if and only if||= 0. We write= 123...|| to denote the components of when|| 6= 0.

    We writef to denote the identity map on AFand, for 0 with || 6= 0,we definef = f1 f2 ... f||.

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    We define, for allm = 0, 1, 2, ...,Vm=

    S{0:||=m}

    f({A,B,C}).

    We also defineV =S

    m=0

    Vm. That is, V is the set of all vertices of the Sierpinski triangle AF.

    We now follow a standard construction due to Kigami, see Strichartz27 . Let u : V R. Then we define, forallm = 0, 1, 2, ...,

    Em(u) = P

    {x,yVm:|xy|=2m}(u(x) u(y))2.

    Then if

    E(u) = limm

    (5

    3)mEm(u)

    exists, we say that "u has finite energy E(u)". We also say that "u is in the domain of (the closure of) theLaplacian", see Strichartz27 . The value of the renormalization constant 5

    3follows from Equation (6) below.

    Suppose we are given the values

    u(A) = hA, u(B) =hB, u(C) = hC. (3)

    Then the corresponding harmonic functionh : V Ris defined to be the function u which minimizesEm(u)foreach m, subject to the constraints in Equation (3).

    We are going to construct explicityh. It is useful, for generalizations, to understand where the various specialnumbers and matrices come from. Let a denote the midpoint of the line segment BC, b denote the midpoint ofthe line segment C A,and c denote the midpoint ofAB . Then V1= {A,B,C,a,b,c}. We have

    E0(u) = |hA hB|2 + |hB hC|2 + |hC hA|2

    and

    E1(u) = |hA u(c)|2

    + |u(c) hB|2

    + |hB u(a)|2

    + |u(a) hC|2

    +|hC u(b)|2 + |u(b) hA|2 + |u(a) u(b)|2 + (|u(b) u(c)|2 + |u(c) u(a))|2.

    We minimizeE1(u)with respect to the values u(a), u(b), u(c).Wefind that at the minimum, where u(a) = h(a),u(b) = h(b),and u(c) = h(c),

    E1(u)

    u(a) = (h(a) hB) + (h(a) hC) + (h(a) h(c)) + (h(a) h(b)) = 0,

    that is,4h(a) h(b) h(c) = hB+ hC

    and two other similar equations which may be obtained by cyclic permutation. It follows that

    4 1 11 4 11 1 4

    h(a)h(b)

    h(c)

    =

    hB+ hChC+ hA

    hA+ hB

    .

    Inverting, we find

    h(a)h(b)h(c)

    =

    3

    10

    1

    10

    1

    101

    10

    3

    10

    1

    101

    10

    1

    10

    3

    10

    0 1 11 0 11 1 0

    hAhBhC

    =

    1

    5

    2

    5

    2

    52

    5

    1

    5

    2

    52

    5

    2

    5

    1

    5

    hAhBhC

    . (4)

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    It follows that

    E1(h) =

    6

    5hAhC

    6

    5hAhB+

    6

    5h2A+

    6

    5h2B

    6

    5hBhC+

    6

    5h2C

    =

    3

    5E0(h).

    Now observe that A= f1(A), B =f2(B), C=f3(C), a= f3(B) =f2(C), b = f1(C) =f3(A),and c = f2(A) =

    f1(B).Using this, we deduce from (4) that

    h(fi(A))h(fi(B))h(fi(C)

    =Ai

    hAhBhC

    ,

    where

    A1=

    1 0 02

    5

    2

    5

    1

    52

    5

    1

    5

    2

    5

    , A2 =

    2

    5

    2

    5

    1

    5

    0 1 01

    5

    2

    5

    2

    5

    , A3=

    2

    5

    1

    5

    2

    51

    5

    2

    5

    2

    5

    0 0 1

    . (5)

    By iterating the above steps, we readily deduce that

    Em(h) =3

    5

    Em1(h) = (3

    5

    )mE0(h), (6)

    and

    h(f12...m(A))h(f12...m(B))h(f12...m(C))

    =Am

    h(f12...m1(A))h(f12...m1(B))h(f12...m1(C))

    =AmAm1 ...A1

    hAhBhC

    ,

    for all 12...m 0 with m 1. Strichartz27 , p.17, says that it should be possible to obtain any desiredinformation about harmonic functions from this last expression but that in practice it may require a lot of work.We are going to show that it is quite easy to obtain a satisfactory description ofh, both theoretical and practical,starting from this expression.

    5 Harmonic functions and the Kigami triangle

    5.1 Values of harmonic functions via the chaos game

    We depart from the standard construction. We letm tend to infinity, use the continuity ofh on AF, andexploit the uniform convergence in (1) to deduce that that A := lim

    mAmAm1 ...A1 exists for all and

    h(F())h(F())h(F()))

    =A

    hAhBhC

    for all .

    We rewrite this as

    h(F()) = Ah0

    where h(F())= (h(F()), h(F()), h(F()))> and h0= (hA, hB, hC)>. This provides the value ofh at thepointF(). In order to obtain global information let v> = (v1, v2, v3) R3 be a test vector and consider

    v.h(F()) = v.Ah0= A> v.h0

    It follows thatA>i A

    > v.h0= A

    >iv.h0= v.h(F(i))for i = 1, 2, 3.

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    Hence we can apply algorithms of the chaos game type to evaluate sequences of values {v.h(xk, yk)}k=0 wherethe sequence of points {(xk, yk)}k=0 are distributed ergodically on the Sierpinski triangle according, say, to thestandard measure. For simplicity suppose we start from a point (x0, y0) = F()and vector v0= A> v. Then atthek th iterative step we choose i {1, 2, 3}randomly, independently of all other choices, and set

    (xk, yk) = fi(xk

    1, yk

    1), vk = A>i vk

    1, v0.h(xk, yk) = vk.h0 fork = 1, 2, 3,... (7)

    Notice that the elements of the columns of the matrices A>i sum to one. Hence each of the transformationsrepresented by the A>i maps the plane

    := {(x1, x2, x3) R3 :x1+ x2+ x3= }

    into itself for each R. It follows that ifv0 1then vk 1for allk, and then the last equation in (7) yields

    h(xk, yk) = (vk,1hA+ vk,2hB+ vk,3hC)

    for k = 1, 2, 3... Thus we have a simple algorithm which allows us to compute, rapidly, values of the harmonicfunctionh.

    5.2 The Kigami triangle

    We can obtain much finer information by looking at the set of points Tv0 ={A> v0: } R3. ClearlyTv0

    obeys the self-referential equation

    Tv0

    =A>1(Tv0) A>2(Tv0) A>3(Tv0).

    Notice that v0 implies Tv0 . Without loss of generality we choose C= 1and we represent the pointsof : 1 R2 by means of the transformation ((, , )) = (, )with inverse 1(, ) = (1 , , ). Wereadily find, in matrix representation,

    = 0 1 0

    0 0 1

    and 1 =

    1 11 0

    0 1

    +

    10

    0

    Then the new transformations gi= A>i 1 : R2 R2 are defined explicitly by

    g1

    =

    2

    5

    1

    51

    5

    2

    5

    (8)

    g2

    =

    3

    5 0

    15

    1

    5

    +

    2

    51

    5

    g3

    =

    1

    5

    15

    0 35

    +

    1

    52

    5

    Notice that each of these affine transformations is strictly contractive. Hence

    G= {R2

    :g1, g2, g3}

    is a hyperbolic IFS. LeteTv0

    =(Tv0

    ). TheneTv0

    =g1(eTv0)g2(eTv0)g3(eTv0). This set is compact and nonempty.Hence it must be the unique attractor of the IFS G , that is

    AG =eTv0 .It is easy to make as sketch ofAG with the aid of the chaos game algorithm. See Figure 1. We refer to AG

    as a Kigami triangle because he first noted the simplicity of the Sierpinski triangle when it is represented using

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    harmonic coordinates. Alexander Teplyaev3 1 2 8 2 9 30 appears to be the first to publish pictures of a Kigamitriangle and some of its relatives,

    We now prove that the Sierpinski triangle AFand the Kigami triangle AG are homeomorphic. Let Ndenotethe triangle with vertices

    eA= (0, 0),

    eB = (1, 0),and

    eC= (0, 1). Let

    ea= ( 2

    5, 2

    5),

    eb= ( 1

    5, 2

    5) and

    ec= ( 2

    5, 1

    5).Then

    we note thatg1(N) g2(N) = ec, g2(N) g3(N) = ea, g3(N) g1(N) = eb.

    It follows thatg1(AG) g2(AG) = ec, g2(AG) g3(AG) = ea, g3(AG) g1(AG) = eb.

    This implies that AFhas the same code structure as AG . Hence, by Theorem 3, the Sierpinski triangle AF andthe Kigami triangle AG are homeomorphic. This is a much shorter proof, for those familiar with IFS theory,than the one given by Teplyaev31 ; in essense the proofs are the same, but the isolation of the tools used is muchclearer in the IFS framework. We will use these same tools to prove that the basic curves, from which the Kigamitriangle may be constructed, are continuously differentiable.

    Let H : AF AG denote the homeomorphism between the attractors. This continuous invertible transfor-mation, with continuous inverse H1 :AG AF,relates each point on one attractor to the corresponding point

    on the other attractor with the same set of addresses. It readily follows that we can write, consistently with ourearlier usage(, ) = H(x, y).

    The harmonic function h : AF Rbecomes, on AG ,eh: AG Rgiven byeh= h H1 andeh(, ) = h(x, y).

    and we have eh(, ) = h(x, y).Finally note that eh(, ) = hA(1 ) + hB+ hC= h(x, y).In the new coordinates, the harmonic function may be described using the plane which passes through thepoints (hA, 0, 0), (0, hB, 0) and (0, 0, hC). Treat the Kigami triangle as being located on the plane = 0, inthree-dimensional space described using rectangular coordinates (, , ) so that a typical point on the Kigamitriangle is represented by (0, , ). If the unique corresponding point on ,with the same ,and coordinates,is (, , )theneh(, ) = . See Figure 1.

    To obtain a symmetrical version of the Kigami triangle we make the change of variable

    T=

    1 1

    2

    0

    3

    2

    , T1 =

    1 1

    3

    3

    0 23

    3

    which corresponds to mapping the vertices(0, 0),(1, 0)and(0, 1)to(0, 0),(1, 0)and(0,

    3/2), respectively, whichlie of an equilateral triangle. Then the linear transformations become self-adjoint, specifically

    eg1() = R 23

    DR23

    (),eg2() = R23

    DR 23

    () + t2,eg3() = D() + t3where

    =

    , D=

    1

    5 0

    0 35

    , =

    , t2=

    1

    21

    10

    3

    , t3=

    2

    51

    5

    3

    andR denotes clockwise rotation about the origin through angle . See Figure 2.

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    Figure 1: The Sierpinski triangle and the Teplyaev triangle. They are homeomorphic because they have the samecode structure. Unlike the Sierpiskis triangle, the Tephaev triangle has a unique "direction" at every point of itsouter boundary and at every point on the internal closed curves.

    Figure 2: An equilateral Kigami triangle.

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    We introduce the vectors

    w1=

    00

    ,w2=

    10

    ,w3=

    1

    23

    2

    ,

    w4=

    3

    23

    5

    ,w5=

    2

    53

    5

    ,w6 =

    1

    23

    10

    .

    Then we note that

    eg1(w1) = w1, eg1(w2) = w6, eg1(w3) = w5, (9)eg2(w2) = w2, eg2(w1) = w6, eg2(w3) = w4,eg3(w3) = w3, eg3(w2) = w4, eg3(w1) = w5.A Kigami triangle can thought of as a type of fractal interpolation, see Barnsley 2 3 , taking the topological form ofa Sierpinski triangle in place a line segment, and passing through, or interpolating, the points w1,w2,w3,w4,w5,and w6.

    5.3 The nature of the Kigami triangle

    By zooming in and looking closely at pictures ofeTv0

    we are led to the impression that there is a well-definedtangent direction at everypoint on each of the curves which define the outer boundary and also on all of theinternal loops, and that this tangent direction depends continuously on location. This is indeed the case. ThusIFS theory provides geometrical information which makes more precise the known fact, proved using Osceledetstheorem on products of random matrices, that there is a well-defined direction atalmost everypoint, with respectto the standard measure. The proof is quite beautiful.

    Theorem 1. The attractoreG of the IFSeG = {R

    2; eg1, eg2}is the graph of a continuously differentiable functionb : [0, 1] [0, 1].

    Proof. It follows from Equations (9) that

    eg1(w1) = w1, eg1(w2) = w6eg2(w2) = w2, eg2(w1) = w6Furthermore eg1(w1w2w3) = w1w6w5 andeg2(w1w2w3) = w6w2w4,and the triangles w1w6w5and w6w2w4are both contained in the triangle w1w2w3. It follows that that the codestructure of the attractor of the IFS {R2; eg1, eg2}is the same as the code structure of the attractor of the IFS

    {R; f1, f2}

    whose attractor is the line segment [0, 1] R. It follows, via the fractal homeomorphism theorem, thateGis thegraph of a continuous function b : [0, 1] [0, 1].

    Finally we outline the proof that b is differentiable and that its derivative is continuous. Consideration of themanner in which the two functions which compriseeG map ellipses into ellipses leads to the conclusion that theslope ofeGexists at every point if and only if it is defined by the attractor of the IFS

    H= {[ 13

    , 1

    3]; , w1(x) =

    3 3x3x 5 , w2(x) =

    3 + 3x3x + 5

    }.

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    That is, we need to show that H is indeed a hyperbolic IFS; then the slope ofeGat the point whose address is is the same as H() provided the address structures the two IFSseG and Hare the same. It is readily verifiedthatH is indeed a hyperbolic IFS. Moreover

    w1([1

    3,

    13

    ]) = [ 13

    , 0]and w2([1

    3,

    13

    ]) = [0, 1

    3].

    It follows that the attractor ofH is the interval [ 13

    , 13

    ] and that the two IFSseG and H do indeed have thesame address structures. Hence, via the fractal homeomorphism theorem,eGis not only differentiable but also itis homeomorphic to the set of its slopes. Hence it is continuously differentiable.

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