Mean value properties on Sierpinski type fractals Hua Qiu (Joint work with Robert S. Strichartz) Department of Mathematics Nanjing University, Cornell University Department of Mathematics, Nanjing University First Previous Next Last 1
1. Introduction For a domain Ω in Euclidean space, a continuous function u is harmonic ( u = 0) if and only if 1 u(y)dy = u(x) B r (x) B r (x) if B r (x) Ω where B r (x) is the ball of radius r about x. Department of Mathematics, Nanjing University First Previous Next Last 2
More generally, if u is not assumed harmonic but u is a continuous function, then 1 1 lim r 0 r 2( u(y)dy u(x)) = c n u(x) (1) B r (x) B r (x) for the appropriate dimensional constant c n. Department of Mathematics, Nanjing University First Previous Next Last 3
What are the fractal analogs of these results? What are the fractal analogs of balls on which we do the averaging? So if K is a fractal and x K, we would like to know that there is a sequence of sets B k (x) containing x with k B k(x) = {x} such that 1 µ(b k (x)) B k (x) for every harmonic function u. u(y)dy = u(x) Department of Mathematics, Nanjing University First Previous Next Last 4
We call B k (x) the k-level mean value neighborhood of x. Let K = SG equipped with the standard harmonic structure and the standard self-similar measure. If x is a nonboundary vertex point, we can easily answer the question with B k (x) being the two union of the k level neighboring cells of x. What about the generic points? Department of Mathematics, Nanjing University First Previous Next Last 5
More generally, for general u not assumed harmonic but belonging to dom, does an analogous formula of (1) still hold? Department of Mathematics, Nanjing University First Previous Next Last 6
2. Mean value neighborhoods on Sierpinski gasket Consider any cell F w (SG) = C w with boundary points F w (q i ) = p i. For any point x C w there exist coefficients a i (x) such that h(x) = a i (x)h(p i ) i for any given harmonic function h. Department of Mathematics, Nanjing University First Previous Next Last 7
Since constants are harmonic we must have a i (x) = 1, and by the maximum principle all a i (x) 0. i We can compute {a i (x)} for x any junction point by using the harmonic extension algorithm. Department of Mathematics, Nanjing University First Previous Next Last 8
Let W denote the triangle in R 3 with boundary points (1, 0, 0), (0, 1, 0), (0, 0, 1). So {a i (x)} W for any x C w. Of course not every point in W occurs in this way. Department of Mathematics, Nanjing University First Previous Next Last 9
Figure 1. Department of Mathematics, Nanjing University First Previous Next Last 10
Given a set B SG, define M B (f) = 1 B fdµ the mean value of function f on B. By linearity, M B (h) = a i h(p i ) i for some coefficients {a i } for any harmonic function h. We also have i a i = 1 by considering h 1. Department of Mathematics, Nanjing University First Previous Next Last 11
Idea: We hope that {a i } W if C w B. If this is true, we have a map {B} W. If we can show that the map is onto for some reasonable class of sets B, then we can get our first B 0 (x) for every x C w, and then by zooming in get a full sequence B k (x). Department of Mathematics, Nanjing University First Previous Next Last 12
Observation: Note that M Cw (h) = i 1 3 h(p i) so we hit the center point ( 1 3, 1 3, 1 3 ) of W by taking B = C w. If we take B to be the union of C w and one of its neighboring cell of the same size, then for the point of intersection. M B (h) = h(p i ) Department of Mathematics, Nanjing University First Previous Next Last 13
So we can get the 3 vertices (1, 0, 0), (0, 1, 0), (0, 0, 1) of W in this way. By varying B continuously between the two we can hit a curve in W joining the center to any vertex. Figure 2. Department of Mathematics, Nanjing University First Previous Next Last 14
This looks like the beginning of an argument to show that the mapping is onto. By the observation, we should require B to be a subset of the union of C w and its 3 neighbors, containing C w. Reasonable choice of B: connected, symmetric, depends only on the relative geometry of x and C w, independent of the size of C w and the location of C w. Department of Mathematics, Nanjing University First Previous Next Last 15
Denote by C 0, C 1, C 2 the three neighboring cells of the same size of C w, intersecting C w at p i for each i. Write D w the union of C w and its 3 neighbors, i.e., D w = C w C 0 C 1 C 2. Consider a set B, C w B D w. B = C w E 0 E 1 E 2, where E i = B C i, i = 0, 1, 2. Department of Mathematics, Nanjing University First Previous Next Last 16
We restrict each E i to be a triangle sub-domain of C i, symmetric under the reflection symmetry that fixes p i, containing p i as one of its boundary points. Figure 3. Department of Mathematics, Nanjing University First Previous Next Last 17
Suppose the height of C w is r, then the height of E i should be c i r for 0 c i 1. Hence we can write the set B = B(c 0, c 1, c 2 ). For example, B(0, 0, 0) = C w and B(1, 1, 1) = D w. Denote by B = {B(c 0, c 1, c 2 ) : 0 c i 1} the family of all such sets. Department of Mathematics, Nanjing University First Previous Next Last 18
Let T w denote the map from B to π W (the plane containing W ) described before. T w can be viewed as a nonlinear vector valued function from [0, 1] 3 to π W. For simplicity, we may write T w ({c i }) = {a i } for each set B(c 0, c 1, c 2 ). Department of Mathematics, Nanjing University First Previous Next Last 19
Fact 1. The map T w is independent of the particular choice of C w. Hence we can drop the subscript w on T for simplicity. The proof of Fact 1 benefits from the symmetric properties of E i s. Department of Mathematics, Nanjing University First Previous Next Last 20
Fact 2. There exists B B, such that T (B) / W. For example, a easy computation shows that T (0, 1, 1) = { 1 9, 5 9, 5 9 }. However, if we can show that the image of the map T can cover the triangle W, things will still go well. Department of Mathematics, Nanjing University First Previous Next Last 21
Theorem 1. The map T from B to π W fills out a region W which contains the triangle W. Sketch of the proof. Step 1. Consider a subfamily B 1 = {B(0, 0, c 2 ) : 0 c 2 1} of B. If we restrict the map T to B 1, by varying c 2 continuously between 0 and 1 we can hit a curve (it is a line segment which follows from the symmetry of E 2 ) in W joining the center O to an vertex point N. Department of Mathematics, Nanjing University First Previous Next Last 22
Step 2. Consider another subfamily B 2 = {B(0, c, c) : 0 c 1} of B. If we restrict the map T to B 2, by varying c continuously between 0 and 1 we can hit a curve (it is a line segment which follows from the symmetric effect of E 1 and E 2 ) in W joining the center O to an point Q. A directly computation shows that Q lies outside of the triangle W. Step 3. T ({B(0, c, 1) : 0 c 1}) is a curve located outside of the triangle W, joining N to Q. The intersection of this curve and W consists exactly only one point N. Department of Mathematics, Nanjing University First Previous Next Last 23
Step 4. Fix a number 0 y 1. Consider a subfamily C y = {B(0, c, y) : 0 c y} of B. If we restrict the map T on C y, by varying c continuously between 0 and y we can hit a curve Γ y joining the two points T (B(0, 0, y)) and T (B(0, y, y)). The first endpoint T (B(0, 0, y)) lies on the line segment ON and the second endpoint T (B(0, y, y)) lies on the line segment OQ. Hence if we vary y continuously between 0 and 1, we can fill out the 1/6 region of W. Then by exploiting the symmetry, we get the desired result. Department of Mathematics, Nanjing University First Previous Next Last 24
Figure 4. Department of Mathematics, Nanjing University First Previous Next Last 25
Let B be a subfamily of B defined by B = {B(0, c 2, c 3 ) : 0 c 2, c 3 1} {B(c 1, 0, c 3 ) : 0 c 1, c 3 1} {B(c 1, c 2, 0) : 0 c 1, c 2 1}, i.e., B consisting of those elements B in B which have the decomposition form B = C w E 1 E 2 or B = C w E 1 E 3, or B = C w E 2 E 3. Department of Mathematics, Nanjing University First Previous Next Last 26
Remark of Theorem 1. Actually, we have proved that W = T (B). Moreover, if we use B in stead of B. Then the map is one-to-one. T : B W Hence for each x C w, there exists a unique set B B, C w B D w, such that M B (h) = h(x) for any harmonic function h. We call the set B associated to this C w a k level mean value neighborhood of x where k is the length of w. Department of Mathematics, Nanjing University First Previous Next Last 27
Given a point x SG \ V 0, let k 0 be the smallest value of k such that there exists a k level cell C w containing x but not intersecting V 0. k 0 depends on the location of x in SG. Then we can find a sequence of words w (k) of length k (k k 0 ) and a sequence of mean value neighborhoods B k (x) associated to C w (k). {B k (x)} k k0 forms a system of neighborhoods of the point x satisfying k k 0 B k (x) = {x}. Department of Mathematics, Nanjing University First Previous Next Last 28
3. Mean value properties of functions in dom Given a point x SG \ V 0, we want to define c B such that M B (u) u(x) c B u(x) for u dom. Let v be a function satisfying v 1. Define c B by c B = M B (v) v(x). Department of Mathematics, Nanjing University First Previous Next Last 29
Note that the value of c B is independent of which v, because any two differ by a harmonic function h and M B (h) h(x) = 0. So we can choose v = SG G(, y)dµ(y). (v vanishes on the boundary of SG, G is the Green s function). Department of Mathematics, Nanjing University First Previous Next Last 30
c B depends only on the relative geometry of B and C w and the size of C w, not on the location of x or C w in SG. Moreover, Theorem 2. There exist two positive constants c 0 and c 1, such that 1 c 0 5 c 1 k B c 1 5 k for any k level mean value neighborhood B. Department of Mathematics, Nanjing University First Previous Next Last 31
Given a point x and C w a k level neighborhood of x, for any u dom 2, we can write u = h (k) + ( u(x))v + R (k) on C w, where h (k) is a harmonic function defined by h (k) + ( u(x))v Cw = u Cw. Department of Mathematics, Nanjing University First Previous Next Last 32
It is not hard to prove that Fact 3. The remainder satisfies on C w (also on B k (x)). R (k) = O(( 3 5 1 5 )k ) (This looks like the Taylor expansion of u at x.) Department of Mathematics, Nanjing University First Previous Next Last 33
Proof of Fact 3. It is easy to check that y R (k) (y) = y u(y) y u(x) and R (k) (y) vanishes on the boundary of C w. Hence R (k) is given by the integral of y u(y) y u(x) on C w against a scaled Green s function. Since the scaling factor is ( 1 5 )k, and y u(y) y u(x) c(3/5) k ( u satisfies the Holder condition with γ = 3 5 ), we get Department of Mathematics, Nanjing University First Previous Next Last 34
the desired result. A more generalized version of Fact 3 is Fact 3 Let u dom with g = u satisfying the following Holder condition g(y) g(x) cγ k, (0 < γ < 1) for all y C w, then the remainder satisfies on C w (also on B k (x)). R (k) = O((γ 1 5 )k ) Department of Mathematics, Nanjing University First Previous Next Last 35
Using the Taylor expansion of u at x and Theorem 2, we have = = 1 c Bk (x) 1 c Bk (x) 1 c Bk (x) (M Bk (x)(u) u(x)) u(x) (M Bk (x)(r (k) ) R (k) (x)) O((γ 1 5 )k ) = O(γ k ). Department of Mathematics, Nanjing University First Previous Next Last 36
Hence we get lim k 1 c Bk (x) (M Bk (x)(u) u(x)) = u(x), which is a fractal analogous formula of (1). Department of Mathematics, Nanjing University First Previous Next Last 37
4. Generalization on a class of Sierpinski type fractals Let K be a p.c.f. self-similar fractal. We assume that V 0 = 3 and all structures possess full D3 symmetry. A) We can choose all c i,j = 1 in defining the graph energy on Γ 0, and all the resistance renormalization factors r i are the same number r. B) We must have µ 1 = µ 2 = µ 3, i.e., µ is the standard self-similar measure on K. Department of Mathematics, Nanjing University First Previous Next Last 38
SG 3, SG n, Hexagasket, 3-dimensional SG... For the first question on how to find the mean value neighborhoods, it is done. However, for the second question on how to estimate the c B constants, it is still on working. Department of Mathematics, Nanjing University First Previous Next Last 39
Thank you! Department of Mathematics, Nanjing University First Previous Next Last 40