DISCRETE MINIMAL ENERGY PROBLEMS

Transcription

1 DISCRETE MINIMAL ENERGY PROBLEMS Lecture II E. B. Saff Center for Constructive Approximation Vanderbilt University University of Crete, Heraklion May, 2017

2 Recall Notation ω N = {x 1,..., x N } A, A compact, infinite K : A A R {+ }, symmetric and l.s.c. E K (ω N ) = i j K (x i, x j ) E K (A, N) = min{e K (ω N ) : ω N A, #ω N = N} W K (A) := min K [A] = min µ M(A) µ M(A) K (x, y) dµ(x)dµ(y) A A µ A equilibrium measure, I K [µ A ] = W K (A) ν(ω N ) = 1 N x ω N δ x, normalized counting measure

3 Recall Notation ω N = {x 1,..., x N } A, A compact, infinite K : A A R {+ }, symmetric and l.s.c. E K (ω N ) = i j K (x i, x j ) E K (A, N) = min{e K (ω N ) : ω N A, #ω N = N} W K (A) := min K [A] = min µ M(A) µ M(A) K (x, y) dµ(x)dµ(y) A A µ A equilibrium measure, I K [µ A ] = W K (A) ν(ω N ) = 1 N x ω N δ x, normalized counting measure Fundamental Thm: lim N E K (A, N)/N 2 = W K (A). Every weak limit measure of the sequence {ν(ωn )} N=2 for N-point K -energy minimizing configurations ωn is an equilibrium measure for the continuous problem on A.

4 Riesz s-energy in Euclidean Space Hereafter A R p and = x y, Euclidean distance. The Riesz s-kernel is defined by K s (x, y) := 1 x y s, s > 0; K 1 log(x, y) := log, x, y A. x y We write E s (ω) := E Ks (ω), E s (A, N) = E Ks (A, N), s > 0 or s = log. For p = 3, s = 1, get Coulomb kernel. For A R p and s = p 2, we get Newton kernel.

5 Some Basic Properties of Riesz Energy A B E s (A, N) E s (B, N) E s (A + x, N) = E s (A, N) E s (αa, N) = α s E s (A, N) E s (A, N) is continuous in s for s > 0 E s (A, N) N(N 1) lim = E log (A, N) s 0 + s and for fixed N, every cluster point of minimal s-energy N-point configurations ω (s) N as s 0+ is an N-point minimal log energy configuration.

6 Best-Packing and Riesz Energy (s ) Recall: Separation distance of ω N = {x 1,..., x N } A δ(ω N ) := N-point best-packing distance on A, min x i x j. 1 i j N δ N (A) := sup{δ(ω N ) : ω N A, #ω N = N}, ω N is best-packing configuration if δ(ω N ) = δ N(A).

7 Best-Packing and Riesz Energy (s ) Recall: Separation distance of ω N = {x 1,..., x N } A δ(ω N ) := N-point best-packing distance on A, min x i x j. 1 i j N δ N (A) := sup{δ(ω N ) : ω N A, #ω N = N}, ω N is best-packing configuration if δ(ω N ) = δ N(A). Proposition (Prove it) For each fixed N 2, lim E s(a, N) 1/s = 1 s δ N (A). Moreover, every cluster configuration as s of s-energy minimizing N-point configurations on A is an N-point best-packing configuration on A.

8 Amusing Exercise Let A = S 1, the unit circle in the plane R 2. Prove that for each s, 0 < s < or s = log the N-th roots of unity (or any rotation of them) are minimal s-energy points for S 1. Moreover, they are the only sets of minimal energy points. More generally: Prove this is true if K (x, y) = f ( x y ), where f is strictly convex and decreasing on (0, 2].

9 S 2 = {x R 3 : x = 1} What about minimal Riesz energy points on S 2 for N = 2?

10 S 2 = {x R 3 : x = 1} What about minimal Riesz energy points on S 2 for N = 2? N = 3?

11 S 2 = {x R 3 : x = 1} What about minimal Riesz energy points on S 2 for N = 2? N = 3? N = 4?

12 Optimality of Tetrahedron for S 2 = {x R 3 : x = 1} More generally, for the d-dimensional sphere S d R d+1 Theorem If K (x, y) = f ( x y 2 ), with f strictly convex and decreasing, then for 2 N d + 2, the only optimal N-point K -energy configurations on S d are the vertices of a regular (N 1)-simplex inscribed in S d centered at 0.

13 Minimal Riesz s-energy for N = 5 on S2

14 Bipyramid appears optimal for s up s 15 Minimal Riesz s-energy for N = 5 on S 2 bipyramid square-base pyramid

15 Minimal Riesz s-energy for N = 5 on S 2 Ratio of s-energy of bipyramid to s-energy of optimal sq-base pyramid s s Melnyk et al (1977) Bipyramid appears optimal for 0 < s < s where s Recently proved by R. Schwartz (over 150 pages + computer assist). Open problem for s > s + ɛ

16 MORALS Optimal Riesz s-energy configurations, in general, depend on s. Rigorous proofs of computational observations can be quite difficult.

17 What do minimal energy points look like" for large N? What about asymptotics of of the minimal energy as N?

18 LET S TALK FOOTBALL (SOCCER)

19 The Soccer Ball Vital Statistics 32 faces = 20 hexagons + 12 pentagons 90 edges 60 vertices

20 The Soccer Ball Vital Statistics 32 faces = 20 hexagons + 12 pentagons 90 edges 60 vertices Has same structure as Carbon 60 (Bucky Ball)

21 The Soccer Ball Vital Statistics 32 faces = 20 hexagons + 12 pentagons 90 edges 60 vertices Has same structure as Carbon 60 (Bucky Ball) Question (important for France) Is it possible to cover (tile) a soccer ball (sphere) with only hexagons?

22 France=Le Hexagon

23 The Soccer Ball Euler s Formula for Convex Polyhedra F = number of faces V = number of vertices E = number of edges Then Examples Cube: =2 Soccer ball: =2 F + V E = 2.

24 Tiling with 5,6, and 7-gons Assume exactly 3 edges from each vertex. F 5 = number of pentagons F 6 = number of hexagons F 7 = number of heptagons Then So 3V = 2E 5F 5 + 6F 6 + 7F 7 = 2E E = 1 2 (5F 5 + 6F 6 + 7F 7 ), V = 1 3 (5F 5 + 6F 6 + 7F 7 ) Now let s apply Euler Formula:

25 Tiling with 5,6, and 7-gons F + V E = 2 (F 5 + F 6 + F 7 ) (5F 5 + 6F 6 + 7F 7 ) 1 2 (5F 5 + 6F 6 + 7F 7 ) = F 5 + 0F F 7 = 2 F 5 = F Corollary If tiling has only hexagons and pentagons, there must be exactly 12 pentagons. You can t tile with hexagons alone!

26 Why is the regular hexagon so important? For the plane, the hexagon ( more precisely, the equilateral triangular lattice) solves the best-packing problem and the best covering problem. Best-Packing Problem in 2 Dimensions Arrange nonoverlapping disks all of the same radius in the plane so as to cover the maximal percentage of area.

27 Why is the regular hexagon so important? For the plane, the hexagon ( more precisely, the equilateral triangular lattice) solves the best-packing problem and the best covering problem. Best-Packing Problem in 2 Dimensions Arrange nonoverlapping disks all of the same radius in the plane so as to cover the maximal percentage of area.

28 Packing disks in 2 dimensions proportion of : area covered balls (r=1) per : unit area π = =.25 2 π = π 12 = =

29 Solved by Fejes Toth and also by...

30 Solved by Fejes Toth and also by...

31 What do minimum energy points for the sphere S 2 look like? Massive high precision experiments were conducted to determine E s (S 2, N) for s = log, s = 1 and N = 2, 3,..., 200. Bad News: There are many local minima of energy that are not global minima. Moreover, the local minima have energies very close to E s (S 2, N). Good News: Equilibrium points try to distribute themselves over a nearly spherical hexagonal net.

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33 Good News: Equilibrium points try to distribute themselves over a nearly spherical hexagonal net. N = 122 electrons in equilibrium (s = 1)

34 Example: 30K minimal s = 3.5-energy points on S2

35 Example: 30K minimal s = 3.5-energy points on S2

36 Bausch, Bowick, et al Grain Boundary Scars and Spherical Crystallography Science, March 2003 Model grain boundaries

37 Spherical crystallography of colloidosomes Spherical crystallography of colloidosomes! Adsorb, say, latex spheres onto lipid bilayer vesicles or water droplets! Useful for encapsulation of flavors and fragrances, drug delivery [H. Aranda-Espinoza e.t al. Science 285, 394 (1999)]!Strength of colloidal armor plating influenced by defects in shell.! For water droplets, surface tension prevents buckling. Colloidosome = colloids of radius a coating water droplet (radius R) -- Weitz Laboratory Ordering on a sphere! a minimum of 12 5-fold disclinations, as in soccer balls and fullerenes -- what happens for R/a >> 1? Confocal image: P. Lipowsky, & A. Bausch

38 S d := {x R d+1 : x = 1} Uniformly distributed spherical configurations A sequence of N-point configurations ω N = {x 1,N, x 2,N,..., x N,N }, N = 1, 2,... is said to be uniformly distributed on the sphere S d if ν(ω N ) σ d, N, where σ d denotes normalized surface area measure on S d. Equivalently, for all f C(S d ), 1 lim N N N f (x i,n ) = f (x) dσ d (x). S d i=1

39 Simple Examples If ω N = {e 2kπi/N } N 1 k=0, the N-th roots of unity, then these configurations are uniformly distributed on the unit circle S 1. The normalized counting measures converge weak to dθ/2π, normalized arclength on the unit circle. Verify that the same is true for the sequence {e λkπi } k=1, whenever λ is an irrational real number.

40 Distributing Points Uniformly on the Sphere Applications: best-packing on sphere viral morphology crystallography electrostatics stable molecules (fullerenes) data sampling (numerical integration, spherical designs) finite normalized tight frames testing radar for aircraft soccer ball designs Minimizing Riesz energy is one method for generating uniformly distributed points.

41 3 Classical Problems on S 2 For s = 1, get Thomson s Problem for Coulomb potential. Answers known for N = 2,..., 6 and 12. For s with N fixed we get the Tammes Problem; that is, the best-packing problem. Answers known for N = 2,..., 14 and N = 24. For s = 0, related to Smale s Problem # 7 for 21 st century: Design polynomial-time algorithm for constructing ω N S 2 such that E log (ω N ) E log ( S 2, N ) C log N, N 2.

42 Asymptotics of E log (S 2, N) Known: Wagner (1989): E log (S 2, N) = (1/2) log(4/e)n 2 (1/2)N log N + O(N) Rakhmanov, Saff, and Zhou (1994): E log (S 2, N) = (1/2) log(4/e)n 2 (1/2)N log N + C N N, where < C N < for N sufficiently large. Bétermin, Sandier (2016) and Sandier, Serfaty (2015) establish that C log := lim N C N exists. Conjecture: Brauchart, Hardin, and Saff (2011): C log = (1/2) log(4/e)+ζ Λ 2 (0) = log 2 2π +3 log 3 Γ(1/3) =

43 Unit Sphere S d := {x R d+1 : x = 1} WHAT S KNOWN ABOUT LARGE N ASYMPTOTICS? For 0 < s < d. The Riesz s-potential of normalized surface area σ d satisfies d+1 U σ Γ( d 2 )Γ(d s) s (x) = I s [σ d ] = Γ( d s+1 2 )Γ(d s/2), x Sd. K s (x, y) = 1 x y s is strictly positive definite on S d. σ d is the unique Riesz s-equilibrium measure. E s (S d, N) lim N N 2 = W s (S d ) = I s [σ d ] <, and any sequence of optimal N-point s-energy configurations is asymptotically uniformly distributed on S d.

44 What About s d? Not difficult to show W s (S d ) = for s d. Moreover, for s > d, there exist C 1, C 2 > 0 such that C 1 N 1+s/d E s (S d, N) C 2 N 1+s/d, for all N 2. Questions: Does lim E s(s d, N)/N 1+s/d N exist? What about the limiting distribution of N-point minimal s-energy configurations?