Fractals and Fractal Dimensions

Transcription

1 Fractals and Fractal Dimensions John A. Rock July 13th, 2009

2 begin with [0,1] remove 1 of length 1/3 remove 2 of length 1/9 remove 4 of length 1/27 remove 8 of length 1/ remove 16 of length 1/243 and in the limit is the Cantor set Figure 1: Construction of the Cantor set Ω CS. Keep track of lengths l n and multiplicities m n at every stage n, but distinguish between the open and closed intervals.

3 [0,1] Ω CS Figure 2: The Cantor string Ω CS : the complement of the Cantor set x1/27 8 x1/81 2 x1/9 1/3 Figure 3: The lengths of the Cantor string. For the Cantor set, the sequence of lengths L of the disjoint open intervals that make up its complement, called the Cantor string Ω CS, are: L = { l n = 3 n each length has multiplicity m n = 2 n 1 }.

4 Definition 1. The geometric zeta function of a fractal string Ω with lengths L is ζ L (s) = where Re s > dim B ( Ω). l s j = j=1 m n ln, s Theorem 2. If a fractal string Ω (a bounded, open set) in [0, 1] is of total length 1 and has an infinite number of lengths in its sequence L, then { } dim B ( Ω) = inf σ R m n ln σ <, where Ω = [0, 1] \ Ω.

5 The Cantor set Ω CS has dimension D = dim B ( Ω) = log 3 2. How s that? ζ L (s) = = = = m n ln s 2 n 1 3 ns 2 n 1 3 ns 3 s s Thus, when is the denominator of the fraction equal to 0? s = 0 1/2 = 3 s s = log 3 1/2 s = log 3 2.

6 Measures! A measure is like a function on sets, but with sensible restrictions. For instance, let V U and U n for n N be some intervals in [0, 1] and let β be a (positive) measure on [0, 1]. Then, β( ) : {intervals in [0,1]} R + 0 { }, β( ) = 0, β(u) 0, β(u) β(v ), and under certain other conditions (such as having the U n disjoint), ( ) β U n = β(u n ).

7 Figure 4: The first few stages in the construction of a mass distribution β on the Cantor set. At each stage, mass is split from the previous stage in ratios of 1/3 on the left and 2/3 on the right. How should we collect the intervals in a meaningful way in order to generate? What did you come up with? Remember that binomial coefficients yield the multiplicities of the masses at every step. They are: ( ) n = k n! k!(n k)!, where k is the number of powers of 2 in the weight, and n is the stage.

8 Figure 5: The solid black blocks correspond to the closed intervals with regularity α(1, 2). Consider the exponent α is the exponent that satisfies U α = µ(u). Equivalently, we have the following definition: Definition 3. The regularity α(u) of a (Borel) measure µ on a subset U [0, 1] with range in [0, ] is α(u) = log µ(u) log U, where = λ( ) is the Lebesgue measure on [0, 1].

9 Figure 6: The solid black blocks correspond to the closed intervals with regularity 1/2. For the intervals U in the family P, the possible mass values are β(u) = 2k 3 with length U = 1 n 3. So, the regularity values are n α(u) = log 2k 3 n log 1 3 n. To exploit the properties of logarithms and in order to come up with families of sequences of lengths, let k 1 be the number of powers of 2 that appear in the mass β(u). Let k 2 be the number of stages it took to get the k 1 powers of 2. Then, let n be the index of the clump of stages with k 2 steps. What do we get?

10 Figure 7: The solid black blocks correspond to the closed intervals with regularity α(1, 2). For the measure σ and the family P of partitions given by the open and closed intervals in the construction of the Cantor set, the regularity values are α = α(k 1, k 2 ) = log (2nk 1 /3nk 2 ) log (1/3 nk 2 ) = 1 k 1 k 2 log 3 2, where k 1 and k 2 are relatively prime non-negative integers such that k 1 < k 2. Thus, ( ) ζ β P (α(k nk2 1, k 2 ), s) = 3 nk2s, nk 1 for Res large enough.

11 Here s the big idea: What are the dimensions associated with the measure β? For each regularity α(k 1, k 2 ) we get a partition zeta function, and these functions have unique real numbers at which they diverge. Then, as with geometric zeta functions, these values yield a dimension associated with the a given regularity. ζ β P (α(k 1, k 2 ), s) = ( ) nk2 3 nk2s, nk 1 So, determine where these functions diverge. Piece of cake!

12 σ dim (supp( ν)) M 1 log (1/2)log Figure 8: Graph of the abscissa of convergence function f for the measure β. The measure β is multifractal. In our setting, this means that the distribution of weight generates a spectrum of dimensions. The abscissa of convergence function f for the measure β has the form ( ) (α 1) (α 1) f(α) = log 3 2 log 3 ( 1 + (α 1) log 3 2 α log 3 2 ) log 3 ( 1 + ) (α 1). log 3 2

13 σ dim (supp( ν)) M 1 log (1/2)log Figure 9: Graph of the abscissa of convergence function f for the measure β. The maximum of f is attained at α = α(1, 2) = 1 (1/2) log 3 2 and this value coincides with the box dimension of the support of the measure β. That is, dim B (supp(β)) = max{ f(α) α = α(k 1, k 2 )} α = log 3 2.

14 Homework. Really. 1. Prove that the total length (or initial length) does not affect the box dimension of a fractal string with sequence of lengths L = {l j } j=1. 2. Prove that for the measure β, the maximum value of f is attained at α = α(1, 2) = 1 (1/2) log 3 2. BIG hint: This was done in my thesis. You can get a copy of my thesis from my website: Click on the familiar picture (not my face). See pages The same function is attained through a different method in K. Falconer s Fractal Geometry (Mathematical Foundations and Applications), 2nd ed., John Wiley, Chichester, 2003.

15 References K. Falconer, Fractal Geometry (Mathematical Foundations and Applications), 2nd ed., John Wiley, Chichester, M. L. Lapidus, J. Lévy Véhel, J. A. Rock, Fractal strings and multifractal zeta functions, submitted to Adv. Math. October (Also, e-print arxiv:math-ph/06105, 2006.) M. L. Lapidus and M. van Frankenhuijsen, Fractal Geometry, Complex Dimensions and Zeta Functions, Springer H-O. Peitgen, H. Jürgens and D. Saupe, Chaos and Fractals, 2nd ed., Springer J. A. Rock, Zeta Functions, Complex Dimensions of Fractal Strings, and Multifractal Analysis of Mass Distributions, thesis, June Website for Mandelbrot set applet: