Internal Diffusion-Limited Erosion

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1 January 18, 2008 Joint work with Yuval Peres

2 Internal Erosion of a Domain 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.

3 Internal Erosion of a Domain 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.

4 Internal Erosion of a Domain 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.

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

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

7 Questions How many sites are eroded?

8 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.

9 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.

10 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 )?

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

12 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]) = m m + n ;

13 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.

14 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?

15 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? Urn model: choose a ball at random, then remove a ball from the other urn. OK Corral Process: Gunfight with m fighters on one side and n on the other. Williams-McIlroy 98, Kingman 99, Kingman-Volkov 03.

16 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

17 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 ( ) 8 1/4 Z (1) 3 where Z is a standard Gaussian. Why n 3/4? For each integer j 1, let X j,y j be independent exponentially distributed random variables with mean j.

18 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 ( ) 8 1/4 Z (1) 3 where Z is a standard Gaussian. Why n 3/4? For each integer j 1, let X j,y j be independent exponentially distributed random variables with mean j. Then P(X j > Y k ) = j j + k

19 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 ( ) 8 1/4 Z (1) 3 where Z is a standard Gaussian. Why n 3/4? For each integer j 1, let X j,y j be independent exponentially distributed random variables with mean j. Then P(X j > Y k ) = j = P([ j,k],[ j,k 1]). j + k

20 Why n 3/4? For each j 1: Replace the edge (j 1,j) by a segment of length Xj ; and Replace the edge ( j,1 j) by a segment of length Yj.

21 Why n 3/4? For each j 1: Replace the edge (j 1,j) by a segment of length Xj ; and Replace the edge ( j,1 j) by a segment of length Yj. Viewing the entire interval [ n,n] as a single rod of length X X n + Y Y n, let the two ends burn continuously at the same constant rate.

22 Why n 3/4? For each j 1: Replace the edge (j 1,j) by a segment of length Xj ; and Replace the edge ( j,1 j) by a segment of length Yj. Viewing the entire interval [ n,n] as a single rod of length X X n + Y Y n, let the two ends burn continuously at the same constant rate. Claim: The order in which segments of the rod finish burning has the same distribution as the order in which sites become eroded.

23 Why n 3/4? For each j 1: Replace the edge (j 1,j) by a segment of length Xj ; and Replace the edge ( j,1 j) by a segment of length Yj. Viewing the entire interval [ n,n] as a single rod of length X X n + Y Y n, let the two ends burn continuously at the same constant rate. Claim: The order in which segments of the rod finish burning has the same distribution as the order in which sites become eroded. This follows from the memoryless property of exponentials: P(X j Y k x X j Y k ) = P(X j x).

24 Why n 3/4? When the origin burns, the remaining rod has length n n L n = X j Y j. j=1 j=1

25 Why n 3/4? When the origin burns, the remaining rod has length n n L n = X j Y j. j=1 j=1 Since n Var( X j j=1 n j=1 Y j ) = 2 n j=1 j 2 = Θ(n 3 )

26 Why n 3/4? When the origin burns, the remaining rod has length n n L n = X j Y j. Since n Var( X j j=1 j=1 n j=1 we have by the Lindeberg CLT j=1 Y j ) = 2 n j=1 j 2 = Θ(n 3 ) where Z is a Gaussian. L n n 3/2 = Z

27 Why n 3/4? When the origin burns, the remaining rod has length n n L n = X j Y j. Since n Var( X j j=1 j=1 n j=1 we have by the Lindeberg CLT where Z is a Gaussian. L n n 3/2 j=1 Y j ) = 2 n j=1 = Z j 2 = Θ(n 3 ) Thus the number of segments remaining in the rod is order Θ( L n ) = Θ(n 3/4 ).

28 Diaconis-Fulton Addition Finite sets A,B Z d. A B = {x 1,...,x k }.

29 Diaconis-Fulton Addition Finite sets A,B Z d. A B = {x 1,...,x k }. To form A + B, let C 0 = A B and C j = C j 1 {y j } where y j is the endpoint of a simple random walk started at x j and stopped on exiting C j 1.

30 Diaconis-Fulton Addition Finite sets A,B Z d. A B = {x 1,...,x k }. To form A + B, let C 0 = A B and C j = C j 1 {y j } where y j is the endpoint of a simple random walk started at x j and stopped on exiting C j 1. Define A + B = C k.

31 Diaconis-Fulton Addition Finite sets A,B Z d. A B = {x 1,...,x k }. To form A + B, let C 0 = A B and C j = C j 1 {y j } where y j is the endpoint of a simple random walk started at x 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 x 1,...,x k.

33 Questions As the lattice spacing goes to zero, is there a scaling limit?

34 Questions As the lattice spacing goes to zero, is there a scaling limit? If so, can we describe the limiting shape?

35 Questions As the lattice spacing goes to zero, is there a scaling limit? If so, can we describe the limiting shape? Not clear how to define dynamics in R d.

36 The Scaling Limit of Diaconis-Fulton Addition Let A,B R d be bounded open sets with A, B having measure zero.

37 The Scaling Limit of Diaconis-Fulton Addition Let A,B R d be bounded open sets with A, B having measure zero. Lattice spacing δ n 0. Write A :: = A δ n Z d.

38 The Scaling Limit of Diaconis-Fulton Addition Let A,B R d be bounded open sets with A, B having measure zero. Lattice spacing δ n 0. Write A :: = A δ n Z d. Theorem (L.-Peres) There exists a deterministic domain D R d such that A :: + B :: D as n

39 The Scaling Limit of Diaconis-Fulton Addition Let A,B R d be bounded open sets with A, B having measure zero. Lattice spacing δ n 0. Write A :: = A δ n Z d. Theorem (L.-Peres) There exists a deterministic domain D R d such that A :: + B :: D as n, in the following sense: for any ε > 0, with probability one D :: ε A :: + B :: D ε:: for all sufficiently large n,

40 The Scaling Limit of Diaconis-Fulton Addition Let A,B R d be bounded open sets with A, B having measure zero. Lattice spacing δ n 0. Write A :: = A δ n Z d. Theorem (L.-Peres) There exists a deterministic domain D R d such that A :: + B :: D as n, in the following sense: for any ε > 0, with probability one D :: ε A :: + B :: D ε:: for all sufficiently large n, where D ε,d ε are the inner and outer ε-neighborhoods of D.

41 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.

42 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.

43 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.

44 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.

46 Diaconis-Fulton sum Divisible sandpile sum

47 Odometer Function u(x) = total mass emitted from x.

48 Odometer Function u(x) = total mass emitted from x. Discrete Laplacian: u(x) = 1 2d u(y) u(x) y x

49 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

50 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.

51 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).

52 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.

53 Free Boundary Problem Unknown function u, unknown domain D = {u > 0}. u 0 u 1 1 A 1 B

54 Free Boundary Problem Unknown function u, unknown domain D = {u > 0}. u 0 u 1 1 A 1 B u( u A + 1 B ) = 0.

55 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.

56 Least Superharmonic Majorant Given A,B Z d, let γ(x) = x 2 g(x,y) g(x,y), y A y B

57 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}.

58 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}.

59 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}. Claim: odometer = s γ.

60 Proof of the claim Claim: odometer = s γ. Let m(x) = amount of mass present at x in the final state.

61 Proof of the claim Claim: odometer = s γ. Let m(x) = amount of mass present at x in the final state. Then u = m 1 A 1 B

62 Proof of the claim Claim: odometer = s γ. Let m(x) = amount of mass present at x in the final state. Then u = m 1 A 1 B 1 1 A 1 B.

63 Proof of the claim Claim: odometer = s γ. Let m(x) = amount of mass present at x in the final state. Then u = m 1 A 1 B 1 1 A 1 B. Since γ = 1 A + 1 B 1 the sum u + γ is superharmonic, so u + γ s.

64 Proof of the claim Claim: odometer = s γ. Let m(x) = amount of mass present at x in the final state. Then u = m 1 A 1 B 1 1 A 1 B. Since γ = 1 A + 1 B 1 the sum u + γ is superharmonic, so u + γ s. Reverse inequality: s γ u is superharmonic on A B and nonnegative outside A B, hence nonnegative inside as well.

65 Defining the Scaling Limit A,B R d bounded open sets such that A, B have measure zero

66 Defining the Scaling Limit A,B R d bounded open sets such that A, B have measure zero Let D = A B {s > γ}

67 Defining the Scaling Limit A,B R d bounded open sets such that A, B have measure zero Let D = A B {s > γ} where Z Z γ(x) = x 2 g(x,y)dy g(x,y)dy A B

68 Defining the Scaling Limit A,B R d bounded open sets such that A, B have measure zero Let D = 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.

69 Defining the Scaling Limit A,B R d bounded open sets such that A, B have measure zero Let D = 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. Odometer: u = s γ.

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

71 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 )

72 The domain D = {s > γ} for two overlapping disks in R 2.

73 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.

74 Main Result Let A,B R d be bounded open sets such that A, B have measure zero.

75 Main Result Let A,B R d be bounded open sets such that A, B have measure zero. Lattice spacing δ n 0.

76 Main Result Let A,B R d be bounded open sets such that A, B have measure zero. Lattice spacing δ n 0. Theorem (L.-Peres) With probability one D n,r n,i n D as n,

77 Main Result Let A,B R d be bounded open sets such that A, B have measure zero. Lattice spacing δ n 0. Theorem (L.-Peres) With probability one where D n,r n,i n D as n, Dn, R n, I n are the smash sums of A δ n Z d and B δ n Z d, computed using divisible sandpile, rotor-router, and Diaconis-Fulton dynamics, respectively.

78 Main Result Let A,B R d be bounded open sets such that A, B have measure zero. Lattice spacing δ n 0. Theorem (L.-Peres) With probability one where D n,r n,i n D as n, Dn, R n, I n are the smash sums of A δ n Z d and B δ n Z d, computed using divisible sandpile, rotor-router, and Diaconis-Fulton dynamics, respectively. D = A B {s > γ}.

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

80 Internal DLA Divisible Sandpile Rotor-Router Model

81 Steps of the Proof convergence of densities convergence of obstacles

82 Steps of the Proof convergence of densities convergence of obstacles convergence of odometer functions

83 Steps of the Proof convergence of densities convergence of obstacles convergence of odometer functions convergence of domains.

84 Multiple Point Sources Fix centers x 1,...,x k R d and λ 1,...,λ k > 0.

85 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,

86 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.

87 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.

88 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.

89 A Quadrature Identity If h is harmonic on δ n Z d, then M t = h(xt j ) j is a martingale for internal DLA, where (X j t ) t 0 is the random walk performed by the j-th particle.

90 A Quadrature Identity If h is harmonic on δ n Z d, then M t = h(xt j ) j is a martingale for internal DLA, where (X j t ) t 0 is the random walk performed by the j-th particle. Optional stopping: E h(x) = EM T = M 0 = x I n k i=1 λ i δn d h(x i ).

91 A Quadrature Identity If h is harmonic on δ n Z d, then M t = h(xt j ) j is a martingale for internal DLA, where (X j t ) t 0 is the random walk performed by the j-th particle. Optional stopping: E h(x) = EM T = M 0 = x I n k i=1 λ i δn d h(x i ). Therefore if I n D, we expect the limiting domain D R d to satisfy Z k h(x)dx = λ i h(x i ). D i=1 for all harmonic functions h on D.

92 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 superharmonic functions h on D.

93 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 superharmonic functions h on D. 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.

94 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 superharmonic functions h on D. 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.

95 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 superharmonic functions h on D. 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. The boundary of B 1... B k lies on an algebraic curve of degree 2k.

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

97 Two Mystery Shapes Cardioid??

The Limiting Shape of Internal DLA with Multiple Sources

The Limiting Shape of Internal DLA with Multiple Sources January 30, 2008 (Mostly) Joint work with Yuval Peres Finite set of points x 1,...,x k Z d. Start with m particles at each site x i. Each particle