Free Boundary Problems Arising from Combinatorial and Probabilistic Growth Models

Transcription

1 Free Boundary Problems Arising from Combinatorial and Probabilistic Growth Models February 15, 2008 Joint work with Yuval Peres

2 Internal DLA with Multiple Sources Finite set of points x 1,...,x k Z d. Start with m particles at each site x i. Each particle performs simple random walk in Z d until reaching an unoccupied site. Get a random set of km occupied sites in Z d. The distribution of this random set does not depend on the order of the walks (Diaconis-Fulton 91).

3 100 point sources arranged on a grid in Z 2. Sources are at the points (50i,50j) for 0 i,j 9. Each source started with 2200 particles.

4 50 point sources arranged at random in a box in Z 2. The sources are iid uniform in the box [0,500] 2. Each source started with 3000 particles.

5 Questions Fix sources x 1,...,x k R d. Run internal DLA on 1 n Zd with n d particles per source. As the lattice spacing goes to zero, is there a scaling limit? If so, can we describe the limiting shape? Lawler-Bramson-Griffeath 92 studied the case k = 1: For a single source, the limiting shape is a ball in R d. Not clear how to define dynamics in R d.

6 Limiting Shape Fix points x 1,...,x k R d and λ 1,...,λ k > 0. Let I n be the random set of occupied sites for internal DLA in the lattice 1 n Zd, if λ i n d particles start at each site x i :: (closest lattice site to x i ). Theorem (L.-Peres) There exists a deterministic domain D R d such that I n D as n in the following sense: for any ε > 0, with probability one D :: ε I n D ε:: for all sufficiently large n, where D ε,d ε are the inner and outer ε-neighborhoods of D.

7 Overlapping Internal DLA Clusters Idea: First let the particles at each source x i perform internal DLA ignoring the particles from the other sources. Get k overlapping internal DLA clusters, each of which is close to a ball. Hard part: How does the shape change when the particles in the overlaps continue walking until they reach unoccupied sites?

8 Two-source internal DLA cluster built from overlapping single-source clusters.

9 Diaconis-Fulton Addition Finite sets A,B Z d. In our application, A and B will be overlapping internal DLA clusters from two different sources. Write A B = {y 1,...,y k }. To form A + B, let C 0 = A B and C j = C j 1 {z j } where z j is the endpoint of a simple random walk started at y j and stopped on exiting C j 1. Define A + B = C k. Abeilan property: the law of A + B does not depend on the ordering of y 1,...,y k.

10 Diaconis-Fulton sum of two squares in Z 2 overlapping in a smaller square.

11 Divisible Sandpile Given A,B Z d, start with mass 2 on each site in A B; and mass 1 on each site in A B A B. At each time step, choose x Z d with mass m(x) > 1, and distribute the excess mass m(x) 1 equally among the 2d neighbors of x. As t, get a limiting region A B Z d of sites with mass 1. Sites in (A B) have fractional mass. Sites outside have zero mass. Abelian property: A B does not depend on the choices.

12 Divisible sandpile sum of two squares in Z 2 overlapping in a smaller square.

13 Diaconis-Fulton sum Divisible sandpile sum

14 Odometer Function u(x) = total mass emitted from x. Discrete Laplacian: u(x) = 1 2d u(y) u(x) y x = mass received mass emitted = 1 1 A (x) 1 B (x), x A B. Boundary condition: u = 0 on (A B). Need additional information to determine the domain A B.

15 Free Boundary Problem Unknown function u, unknown domain D = {u > 0}. u 0 Alternative formulation: u 1 1 A 1 B u( u A + 1 B ) = 0. u = 1 1 A 1 B on D; u = u = 0 on D.

16 Least Superharmonic Majorant Given A,B Z d, let γ(x) = x 2 g(x,y) g(x,y), y A y B where g is the Green s function for simple random walk g(x,y) = E x #{k X k = y}. Let s(x) = inf{φ(x) φ is superharmonic on Z d and φ γ}. Then odometer = s γ.

17 Defining the Scaling Limit A,B R d bounded open sets such that A, B have measure zero We define the smash sum of A and B as A B = A B {s > γ} where Z Z γ(x) = x 2 g(x,y)dy g(x,y)dy A B and s(x) = inf{φ(x) φ is continuous, superharmonic, and φ γ} is the least superharmonic majorant of γ.

18 Obstacle for two overlapping disks A and B: Z Z γ(x) = x 2 g(x,y)dy g(x,y)dy A B Obstacle for two point sources x 1 and x 2 : γ(x) = x 2 g(x,x 1 ) g(x,x 2 )

19 The domain D = {s > γ} for two overlapping disks in R 2. The boundary D is given by the algebraic curve ( x 2 + y 2) 2 2r 2 ( x 2 + y 2) 2(x 2 y 2 ) = 0.

20 Main Result Let A,B R d be bounded open sets such that A, B have measure zero. Theorem (L.-Peres) With probability one where D n,r n,i n A B as n, Dn, R n, I n are the smash sums of A 1 n Zd and B 1 n Zd, computed using divisible sandpile, rotor-router, and Diaconis-Fulton dynamics, respectively. A B = A B {s > γ}. Convergence is in the sense of ε-neighborhoods: for all ε > 0 (A B) :: ε D n,r n,i n (A B) ε:: for all sufficiently large n.

21 Internal DLA Divisible Sandpile Rotor-Router Model

22 Steps of the Proof convergence of densities convergence of obstacles convergence of majorants convergence of domains.

23 Convergence of Obstacles Green s function estimates: The Green s functions g n on 1 n Zd and g on R d are related by: g n (x,y) = n d 2 g 1 (nx,ny) = g(x,y) + O(n 2 x y d ). Get γ n γ uniformly on compact sets in R d. Next, want to show s n s uniformly on compact sets in R d, where s,s n are the least superharmonic majorants of γ,γ n.

24 Convergence of Majorants Taylor s theorem: 1 n Z d s = 2d R d s + O(n 1 s ). Mollify s and restrict to 1 n Zd to get a function f n with s f n < ε and f n < δ. Since γ n γ s, the function φ n = f n δ x 2 + 2ε is superharmonic and γ n, hence s n. Hence s f n ε φ n 3ε s n 3ε.

25 Convergence of Majorants Other direction: given s n superharmonic on 1 n Zd, how to produce a superharmonic function on R d that is close to s n? Step function s n (constant on cubes) is too crude. Instead, construct a function whose Laplacian is ( s n ) : Z φ n (x) = g(x,y)( s n ) (y)dy. R d Need to truncate to get the integral to converge.

26 Convergence of Domains Let u n = s n γ n (odometer function), and let D n = {u n > 0} be the set of fully occupied sites for the divisible sandpile. Want to show for sufficiently large n, where (A B) :: ε D n (A B) ε:: A B = A B {s > γ}. But converging functions converging supports!

27 A Lemma of Caffarelli Quadratic Growth Lemma: If x / (A B) ε and x D n, there exists y 1 n Zd such that x y ε + 1 n and u n (y) ε 2. Proof (Caffarelli): Let B = B(x,ε). Since u n = 1 1 A 1 B = 1 in B D n the function w(y) = u n (y) x y 2 is harmonic in B D n, so attains its max on the boundary. Since w(x) 0 and w < 0 on D n, the max is not attained on D n, so it must be attained on B. Get that if x D n, then s(y) > γ(y) for some y B(x,2ε), hence y A B and x (A B) 2ε.

28 A Converse to Quadratic Growth Lemma: u : R d R, u 0, u(0) = 0. If u λ, then u(x) c d λ x 2 for a constant c d depending only on d.

29 Adapting the Proof for Rotors Rotor-router odometer: u(x) = total number of particles emitted from x. Can try to control u, using the fact that each site emits approximately equally many particles to each of its 2d neighbors. Instead of u = 1 1 A 1 B, we only know (in Z 2 ) 2 u + 1 A + 1 B 4.

30 Smoothing To do better, let S k u(x) = 1 (2k + 1) 2 u(y) y B k (x) where B k (x) is a box of side length 2k + 1 centered at x. Using = div grad, we get 1 u(z) u(y) S k u(x) = (2k + 1) 2 (y,z) B 4 k (x) ( ) 1 = S k (1 1 A 1 B ) + O k if all sites in S k (x) are occupied.

31 Multiple Point Sources Fix centers x 1,...,x k R d and λ 1,...,λ k > 0. Theorem (L.-Peres) With probability one D n,r n,i n D as n, where Dn, R n, I n are the domains of occupied sites δ n Z d, if λ i δ d n particles start at each site x i :: and perform divisible sandpile, rotor-router, and Diaconis-Fulton dynamics, respectively. D is the smash sum of the balls B(x i,r i ), where λ i = ω d ri d. Follows from the main result and the case of a single point source.

32 A Quadrature Identity Given A,B R d, if h C 1( A B ) is harmonic in A B, then by Green s theorem, Z Z Z ( h u = u h + h u n h ) n u = 0. A B A B (A B) since u and u vanish on (A B). Recalling that u = 1 1 A 1 B, we obtain the identity Z Z Z h dx = h dx + h dx. A B A B Here dx is Lebesgue measure in R d. Holds with inequality for all integrable superharmonic functions h on A B.

33 Quadrature Domains Given x 1,...x k R d and λ 1,...,λ k > 0. D R d is called a quadrature domain for the data (x i,λ i ) if Z D h(x)dx k i=1 λ i h(x i ). for all integrable superharmonic functions h on D. (Aharonov-Shapiro 76, Gustafsson, Sakai,...) The smash sum B 1... B k is such a domain, where B i is the ball of volume λ i centered at x i. Generalizes the mean value property of superharmonic functions, which corresponds to the case k = 1. In two dimensions, the boundary of B 1... B k lies on an algebraic curve of degree 2k.

34 ZZ D h(x,y)dx dy = h( 1,0) + h(1,0)

35 Hele-Shaw Flow Blob of incompressible fluid sandwiched in the narrow gap between two plates. Given x 1,...x k R 2 and λ 1,...,λ k > 0. Identifying one of the plates with R 2, inject volume λ i of fluid at x i for each i. The resulting shape of the blob is the smash sum of disks B 1... B k of area λ i centered at x i. Stamping problem: Starting with blob A B, stamp the two plates together on A B. The resulting shape of the blob will be the smash sum A B.

36 Internal Erosion, or Negative Sources Given a finite set A Z d containing the origin. Start a simple random walk at the origin. Stop the walk when it reaches a site x A adjacent to the complement of A. Let e(a) = A {x}. We say that x is eroded from A. Iterate this operation until the origin is eroded. The resulting random set is called the internal erosion of A.

37 Internal erosion of a disk of radius 250 in Z 2.

38 Internal erosion of a box of side length 500 in Z 2.

39 Questions How many sites are eroded? If A is (say) a square of side length n in Z 2, we would guess that E#eroded sites = Θ(n α ) for some 1 < α < 2. Analogy with diffusion-limited aggregation. What is the probability that a given site x is eroded? For some sites, is this probability o(n α 2 )?

40 Probability of a given site being eroded from a box in Z 2.

41 Internal Erosion in One Dimension Interval A = [ m,n] Z with m 0 n. Transition probabilities are given by gambler s ruin: P([ m,n],[ m,n 1]) = P([ m,n],[1 m,n]) = m m + n ; n m + n. How large is the remaining interval when the origin gets eroded? OK Corral Process: Gunfight with m fighters on one side and n on the other. Williams-McIlroy 98, Kingman 99, Kingman-Volkov 03.

42 The Number of Surviving Gunners Let m = n (an equal gunfight). Theorem (Kingman-Volkov 03) Starting from the interval [ n,n], let R(n) be the number of sites remaining when the origin is eroded. Then as n R(n) = n3/4 where Z is a standard Gaussian. ( ) 8 1/4 Z (1) 3

43 Two Mystery Shapes