Geometry and the Dirichlet Problem in Any Co-dimension

Transcription

1 Geometry and the Dirichlet Problem in Any Co-dimension Max Engelstein (joint work with G. David (U. Paris Sud) and S. Mayboroda (U. Minn.)) Massachusetts Institute of Technology January 10, 2019 This research was partially supported by an NSF Postdoctoral Research Fellowship, DMS and David Jerison s NSF DMS

2 Overview Harmonic Measure: where does a random walk first exit a domain? 2

3 Overview Harmonic Measure: where does a random walk first exit a domain? Figure: A random walk exiting a domain (figure credit Matthew Badger) 2

4 Overview Harmonic Measure: where does a random walk first exit a domain? Figure: A random walk exiting a domain (figure credit Matthew Badger) Dirichlet problem: equilibrium after diffusion. 2

5 Overview Harmonic Measure: where does a random walk first exit a domain? Figure: A random walk exiting a domain (figure credit Matthew Badger) Dirichlet problem: equilibrium after diffusion. Not surprising: a nasty domain can have hidden parts of the boundary 2

6 Overview Harmonic Measure: where does a random walk first exit a domain? Figure: A random walk exiting a domain (figure credit Matthew Badger) Dirichlet problem: equilibrium after diffusion. Not surprising: a nasty domain can have hidden parts of the boundary Surprising: if nothing is hidden, the domain is nice. 2

7 Overview Harmonic Measure: where does a random walk first exit a domain? Figure: A random walk exiting a domain (figure credit Matthew Badger) Dirichlet problem: equilibrium after diffusion. Not surprising: a nasty domain can have hidden parts of the boundary Surprising: if nothing is hidden, the domain is nice. Very surprising (and recent): higher co-dimension analogues. 2

8 Random Walk and Hitting Measure Walker goes to each neighbor with equal probability. Figure: Each neighbor has probability 1/4 3

9 Random Walk and Hitting Measure Walker goes to each neighbor with equal probability. Figure: We keep going until we hit the boundary 3

10 Random Walk and Hitting Measure Walker goes to each neighbor with equal probability. Figure: Backtracking is allowed 3

11 Random Walk and Hitting Measure Walker goes to each neighbor with equal probability. Figure: Stop when we hit the boundary 3

12 Random Walk and Hitting Measure Walker goes to each neighbor with equal probability. Figure: Stop when we hit the boundary 3

13 Random Walk and Hitting Measure Walker goes to each neighbor with equal probability. Hitting Measure: probability RW hits that part of the boundary. Figure: Hitting Measure of Green starting at Red ω Pole (Target) 3

14 Random Walk and Hitting Measure Walker goes to each neighbor with equal probability. Hitting Measure: probability RW hits that part of the boundary. Figure: Hitting Measure Depends on the Pole! 3

15 Random Walk and Hitting Measure Walker goes to each neighbor with equal probability. Hitting Measure: probability RW hits that part of the boundary. Figure: Hitting Measure Depends on the Pole! Hard to compute!!! (Solve 10 eqns with 10 unknowns) Average of neighbors! 3

16 Discrete Harmonic Functions Mean Value Property u(point) = 1 #Neighbors Neighbors u(neighbor). 4

17 Discrete Harmonic Functions Mean Value Property Example: u(point) = 1 #Neighbors Neighbors u(neighbor). Figure: Each red value is the average of the neighboring values 4

18 Discrete Harmonic Functions Mean Value Property Example: u(point) = 1 #Neighbors Neighbors u(neighbor). Figure: Each red value is the average of the neighboring values Uniqueness: Maximum principle! 4

19 Expectation and the Dirichlet Problem Recall: u(point) = 1 #Neighbors Neighbors u(neighbor). Figure: How do we fill in the boundary? 5

20 Expectation and the Dirichlet Problem Recall: u(point) = 1 #Neighbors Neighbors u(neighbor). This is the Dirichlet problem. Figure: How do we fill in the boundary? 5

21 Expectation and the Dirichlet Problem Recall: u(point) = 1 #Neighbors Neighbors u(neighbor). Figure: How do we fill in the boundary? This is the Dirichlet problem. u(point) = u(boundarypoint)ω Point (BoundaryPoint). BoundaryPoints 5

22 Expectation and the Dirichlet Problem Recall: u(point) = 1 #Neighbors Neighbors u(neighbor). This is the Dirichlet problem. Figure: How do we fill in the boundary? u(point) = u(boundarypoint)ω Point (BoundaryPoint). Expected value! BoundaryPoints 5

23 Harmonic Functions in the Continuum Mean Value Property: for all X R n and R > 0, ˆ 1 u(y )dy = u(x ). B(X, R) B(X,R) 6

24 Harmonic Functions in the Continuum Mean Value Property: for all X R n and R > 0, ˆ 1 u(y )dy = u(x ). B(X, R) B(X,R) u satisfies mean value property u = 0 n i=1 2 x i x i u = 0. 6

25 Harmonic Functions in the Continuum Mean Value Property: for all X R n and R > 0, ˆ 1 u(y )dy = u(x ). B(X, R) B(X,R) u satisfies mean value property u = 0 n i=1 2 x i x i u = 0. EX : u(x, y) x 2 y 2 Figure: Harmonic functions don t have local extrema credit: laussy.org 6

26 Harmonic Functions in the Continuum Mean Value Property: for all X R n and R > 0, ˆ 1 u(y )dy = u(x ). B(X, R) B(X,R) u satisfies mean value property u = 0 n i=1 2 x i x i u = 0. EX : u(x, y) x 2 y 2 Figure: Harmonic functions don t have local extrema credit: laussy.org Represents Equilibrium after diffusion. 6

27 Harmonic Measure in the Continuum X Ω R n. E Ω. ω X (E) = Probability a B.M. exits Ω first in E. 7

28 Harmonic Measure in the Continuum X Ω R n. E Ω. ω X (E) = Probability a B.M. exits Ω first in E. (D) = { uf =0 x Ω u f (Q) =f (Q) Q Ω Figure: Brownian Motion exiting a domain (figure credit Matthew Badger) 7

29 Harmonic Measure in the Continuum X Ω R n. E Ω. ω X (E) = Probability a B.M. exits Ω first in E. { uf =0 x Ω (D) = u f (Q) =f (Q) Q Ω ˆ u f (X ) = fdω X. Ω Figure: Brownian Motion exiting a domain (figure credit Matthew Badger) 7

30 Harmonic Measure in the Continuum X Ω R n. E Ω. ω X (E) = Probability a B.M. exits Ω first in E. (D) = { uf =0 x Ω u f (Q) =f (Q) Q Ω Figure: Brownian Motion exiting a domain (figure credit Matthew Badger) ˆ u f (X ) = Ω fdω X. What is the temperature in the interior, given the temperature on the edge? 7

31 When does ω X not see large sets? Bad geometry: ω doesn t see sets of large length. 8

32 When does ω X not see large sets? Bad geometry: ω doesn t see sets of large length. Connectivity. Figure: ω Pole cannot see the other component 8

33 When does ω X not see large sets? Bad geometry: ω doesn t see sets of large length. Connectivity. Figure: Brownian motion cannot go down hallways 8

34 When does ω X not see large sets? Bad geometry: ω doesn t see sets of large length. Connectivity. Cusps Figure: Brownian motion cannot go down hallways Figure: Brownian motion cannot get to the cusp 8

35 So what if ω X doesn t see large sets? (Quantitative) Well posedness of Dirichlet problem: { uf =0 x Ω (D) = u f (Q) =f (Q) Q Ω 9

36 So what if ω X doesn t see large sets? (Quantitative) Well posedness of Dirichlet problem: { uf =0 x Ω (D) = u f (Q) =f (Q) Q Ω ω X = k X dσ (σ is length on Ω). 9

37 So what if ω X doesn t see large sets? (Quantitative) Well posedness of Dirichlet problem: { uf =0 x Ω (D) = u f (Q) =f (Q) Q Ω ω X = k X dσ (σ is length on Ω). k X small: big f can give small u. Vice versa if k X big. 9

38 So what if ω X doesn t see large sets? (Quantitative) Well posedness of Dirichlet problem: { uf =0 x Ω (D) = u f (Q) =f (Q) Q Ω ω X = k X dσ (σ is length on Ω). k X small: big f can give small u. Vice versa if k X big. Figure: Changing data on the cusp doesn t change the solution 9

39 So what if ω X doesn t see large sets? (Quantitative) Well posedness of Dirichlet problem: { uf =0 x Ω (D) = u f (Q) =f (Q) Q Ω ω X = k X dσ (σ is length on Ω). k X small: big f can give small u. Vice versa if k X big. Figure: Changing data on the cusp doesn t change the solution Theorem : D is (quantitatively) well posed iff k X is not too large or too small too often. 9

40 So what if ω X doesn t see large sets? (Quantitative) Well posedness of Dirichlet problem: { uf =0 x Ω (D) = u f (Q) =f (Q) Q Ω ω X = k X dσ (σ is length on Ω). k X small: big f can give small u. Vice versa if k X big. Figure: Changing data on the cusp doesn t change the solution Theorem : D is (quantitatively) well posed iff k X is not too large or too small too often. Call this A 9

41 Three Examples: ω vs Length σ is the length measure. 10

42 Three Examples: ω vs Length σ is the length measure. A disk, Lipschitz domain and Snowflake (figure from Matthew Badger) 10

43 Three Examples: ω vs Length σ is the length measure. For the disk, ω 0 = A disk, Lipschitz domain and Snowflake (figure from Matthew Badger) σ 2πr. 10

44 Three Examples: ω vs Length σ is the length measure. For the disk, ω 0 = A disk, Lipschitz domain and Snowflake (figure from Matthew Badger) σ 2πr. For a Lipschitz domain, if ω 0 (E) = kσ and k is not too small or too big too often. 10

45 Three Examples: ω vs Length σ is the length measure. A disk, Lipschitz domain and Snowflake (figure from Matthew Badger) For the disk, ω 0 = σ 2πr. For a Lipschitz domain, if ω 0 (E) = kσ and k is not too small or too big too often. For the snowflake ω 0 = kσ and k = + or 0 at every point. 10

46 Relationship between Geometry and H.M. Long Corridors, Cusps are problems. 11

47 Relationship between Geometry and H.M. Long Corridors, Cusps are problems. They aren t the only problem! Fractals! Q1 (direct): If Ω is nice does that mean ω X = k X dσ for nice k X? 11

48 Relationship between Geometry and H.M. Long Corridors, Cusps are problems. They aren t the only problem! Fractals! Q1 (direct): If Ω is nice does that mean ω X = k X dσ for nice k X? Q2 (free boundary): If ω X = k X dσ for nice k X, does that mean Ω must be nice? 11

49 Relationship between Geometry and H.M. Long Corridors, Cusps are problems. They aren t the only problem! Fractals! Q1 (direct): If Ω is nice does that mean ω X = k X dσ for nice k X? Q2 (free boundary): If ω X = k X dσ for nice k X, does that mean Ω must be nice? Lots of work: Ahlfors, Bishop, Carleson, David, E., Fabes, Garnett, Hofmann, Jerison, Kenig, Laurentiev, Mayboroda, Nyström, Øksendal, Pipher, Riesz (x2), Salsa, Toro, Uriarte-Tuero, Volberg, Wolff, Zhao...Many More. 11

50 Relationship between Geometry and H.M. Long Corridors, Cusps are problems. They aren t the only problem! Fractals! Q1 (direct): If Ω is nice does that mean ω X = k X dσ for nice k X? Q2 (free boundary): If ω X = k X dσ for nice k X, does that mean Ω must be nice? Lots of work: Ahlfors, Bishop, Carleson, David, E., Fabes, Garnett, Hofmann, Jerison, Kenig, Laurentiev, Mayboroda, Nyström, Øksendal, Pipher, Riesz (x2), Salsa, Toro, Uriarte-Tuero, Volberg, Wolff, Zhao...Many More. When n = 2: connections to the complex analysis. 11

51 Green Function Associated with Harmonic measure ω X is Green function G(X, ). 12

52 Green Function Associated with Harmonic measure ω X is Green function G(X, ). G(X, Y ) >0 X Y Ω G(X, Q) =0 Q Ω Y G(X, Y ) =δ X (Y ) Y Ω. Figure: George Green (figure credit wikipedia) 12

53 Green Function Associated with Harmonic measure ω X is Green function G(X, ). G(X, Y ) >0 X Y Ω G(X, Q) =0 Q Ω Y G(X, Y ) =δ X (Y ) Y Ω. ˆ Ω ˆ ϕ(y )G(X, Y )dy = ϕ(y ) + ϕ(q)dω X (Q). Ω Figure: George Green (figure credit wikipedia) 12

54 Green Function Associated with Harmonic measure ω X is Green function G(X, ). G(X, Y ) >0 X Y Ω G(X, Q) =0 Q Ω Y G(X, Y ) =δ X (Y ) Y Ω. Figure: George Green (figure credit wikipedia) ˆ Ω ˆ ϕ(y )G(X, Y )dy = ϕ(y ) + ϕ(q)dω X (Q). Ω Probability: G(X, Y ) = how likely does B.M. go from X to Y without leaving Ω (tricky!) 12

55 Green Function Associated with Harmonic measure ω X is Green function G(X, ). G(X, Y ) >0 X Y Ω G(X, Q) =0 Q Ω Y G(X, Y ) =δ X (Y ) Y Ω. Figure: George Green (figure credit wikipedia) ˆ Ω ˆ ϕ(y )G(X, Y )dy = ϕ(y ) + ϕ(q)dω X (Q). Ω Probability: G(X, Y ) = how likely does B.M. go from X to Y without leaving Ω (tricky!) Really hard to compute! Known only for a few domains (half plane, disc, polygons...) 12

56 Poisson Kernel dω X (Q) = k X (Q)dσ(Q). 13

57 Poisson Kernel dω X (Q) = k X (Q)dσ(Q). k X is Poisson kernel. 13

58 Poisson Kernel dω X (Q) = k X (Q)dσ(Q). k X is Poisson kernel. Figure: Siméon Denis Poisson (figure credit wikipedia) 13

59 Poisson Kernel dω X (Q) = k X (Q)dσ(Q). k X is Poisson kernel. Figure: Siméon Denis Poisson (figure credit wikipedia) k X = n G(X, ). 13

60 Poisson Kernel dω X (Q) = k X (Q)dσ(Q). k X is Poisson kernel. Figure: Siméon Denis Poisson (figure credit wikipedia) k X = n G(X, ). Green function hard to compute Poisson kernel hard to compute. 13

61 Poisson Kernel dω X (Q) = k X (Q)dσ(Q). k X is Poisson kernel. Figure: Siméon Denis Poisson (figure credit wikipedia) k X = n G(X, ). Green function hard to compute Poisson kernel hard to compute. Q2 Above: If k X is nice does that mean Ω must be nice? 13

62 Poisson Kernel dω X (Q) = k X (Q)dσ(Q). k X is Poisson kernel. Figure: Siméon Denis Poisson (figure credit wikipedia) k X = n G(X, ). Green function hard to compute Poisson kernel hard to compute. Q2 Above: If k X is nice does that mean Ω must be nice? Overdetermined: k X Neumann conditions, Q Ω G(X, Q) = 0, Dirichlet conditions. 13

63 Poisson Kernel dω X (Q) = k X (Q)dσ(Q). k X is Poisson kernel. Figure: Siméon Denis Poisson (figure credit wikipedia) k X = n G(X, ). Green function hard to compute Poisson kernel hard to compute. Q2 Above: If k X is nice does that mean Ω must be nice? Overdetermined: k X Neumann conditions, Q Ω G(X, Q) = 0, Dirichlet conditions. Free Boundary Problem! 13

64 Back to the Questions Q1 (direct): If Ω is nice does that mean ω = kdσ for nice k? Q2 (free boundary): If ω = kdσ for nice k, does that mean Ω is nice? 14

65 Back to the Questions Q1 (direct): If Ω is nice does that mean ω = kdσ for nice k? Q2 (free boundary): If ω = kdσ for nice k, does that mean Ω is nice? Yes! For essentially every value of nice! 14

66 Back to the Questions Q1 (direct): If Ω is nice does that mean ω = kdσ for nice k? Q2 (free boundary): If ω = kdσ for nice k, does that mean Ω is nice? Yes! For essentially every value of nice! Exercise: k = constant iff Ω = B(X, R). 14

67 Back to the Questions Q1 (direct): If Ω is nice does that mean ω = kdσ for nice k? Q2 (free boundary): If ω = kdσ for nice k, does that mean Ω is nice? Yes! For essentially every value of nice! Exercise: k = constant iff Ω = B(X, R). Harder: k C k,α iff Ω C k+1,α (need α (0, 1), Jerison-Kenig!) 14

68 Back to the Questions Q1 (direct): If Ω is nice does that mean ω = kdσ for nice k? Q2 (free boundary): If ω = kdσ for nice k, does that mean Ω is nice? Yes! For essentially every value of nice! Exercise: k = constant iff Ω = B(X, R). Harder: k C k,α iff Ω C k+1,α (need α (0, 1), Jerison-Kenig!) Kenig-Toro 90s, 00s: osc k controls osc Ω. Vice Versa! 14

69 Back to the Questions Q1 (direct): If Ω is nice does that mean ω = kdσ for nice k? Q2 (free boundary): If ω = kdσ for nice k, does that mean Ω is nice? Yes! For essentially every value of nice! Exercise: k = constant iff Ω = B(X, R). Harder: k C k,α iff Ω C k+1,α (need α (0, 1), Jerison-Kenig!) Kenig-Toro 90s, 00s: osc k controls osc Ω. Vice Versa! Hofmann-Martell & Azzam-Mourgoglou-Tolsa 2018: k isn t too small or too big too often (A -condition) iff Ω looks flat at most points and scales (uniformly rectifiable). 14

70 Back to the Questions Q1 (direct): If Ω is nice does that mean ω = kdσ for nice k? Q2 (free boundary): If ω = kdσ for nice k, does that mean Ω is nice? Yes! For essentially every value of nice! Exercise: k = constant iff Ω = B(X, R). Harder: k C k,α iff Ω C k+1,α (need α (0, 1), Jerison-Kenig!) Kenig-Toro 90s, 00s: osc k controls osc Ω. Vice Versa! Hofmann-Martell & Azzam-Mourgoglou-Tolsa 2018: k isn t too small or too big too often (A -condition) iff Ω looks flat at most points and scales (uniformly rectifiable). Takeaway: Geometry of a set is characterized by solutions of Laplacian in complement of the set! 14

71 Digression on Dimension Famous Open Q: What is the (maximal) dimension of the support of ω? 15

72 Digression on Dimension Famous Open Q: What is the (maximal) dimension of the support of ω? Figure: Can harmonic measure live on the whole boundary? (figure credit wikipedia) 15

73 Digression on Dimension Famous Open Q: What is the (maximal) dimension of the support of ω? Figure: Can harmonic measure live on the whole boundary? (figure credit wikipedia) Makarov 85 Jones-Wolff 88: in n = 2, dim ω = 1. 15

74 Digression on Dimension Famous Open Q: What is the (maximal) dimension of the support of ω? Figure: Can harmonic measure live on the whole boundary? (figure credit wikipedia) Makarov 85 Jones-Wolff 88: in n = 2, dim ω = 1. Bourgain 87: in R n, dim ω < n. 15

75 Digression on Dimension Famous Open Q: What is the (maximal) dimension of the support of ω? Figure: Can harmonic measure live on the whole boundary? (figure credit wikipedia) Makarov 85 Jones-Wolff 88: in n = 2, dim ω = 1. Bourgain 87: in R n, dim ω < n. Wolff 95: in R n, n 3 dim ω > n 1, 15

76 Digression on Dimension Famous Open Q: What is the (maximal) dimension of the support of ω? Figure: Can harmonic measure live on the whole boundary? (figure credit wikipedia) Makarov 85 Jones-Wolff 88: in n = 2, dim ω = 1. Bourgain 87: in R n, dim ω < n. Wolff 95: in R n, n 3 dim ω > n 1, Precise value completely open! 15

77 Higher Co-dimension Would like to characterize geometry of higher-co-dimension sets! 16

78 Higher Co-dimension Would like to characterize geometry of higher-co-dimension sets! Think: curve in R 3. 16

79 Higher Co-dimension Would like to characterize geometry of higher-co-dimension sets! Think: curve in R 3. More exotic: snowflake in R 3! 16

80 Higher Co-dimension Would like to characterize geometry of higher-co-dimension sets! Think: curve in R 3. More exotic: snowflake in R 3! Problem: Elliptic PDE don t see sets of co-dim > 2! (removable!) 16

81 Higher Co-dimension Would like to characterize geometry of higher-co-dimension sets! Think: curve in R 3. More exotic: snowflake in R 3! Problem: Elliptic PDE don t see sets of co-dim > 2! (removable!) Why do this? It is fun! 16

82 Higher Co-dimension Would like to characterize geometry of higher-co-dimension sets! Think: curve in R 3. More exotic: snowflake in R 3! Problem: Elliptic PDE don t see sets of co-dim > 2! (removable!) Why do this? It is fun! Applications to Biology? Figure: DNA Straightens and Curls up to Attract/Avoid Enzymes 16

83 Degenerate Elliptic PDE Need degenerate elliptic PDE. 17

84 Degenerate Elliptic PDE Need degenerate elliptic PDE. Degenerate coefficients attract Brownian motion. 17

85 Degenerate Elliptic PDE Need degenerate elliptic PDE. Degenerate coefficients attract Brownian motion. E = (x, φ(x)) R n. φ : R d R n d. 17

86 Degenerate Elliptic PDE Need degenerate elliptic PDE. Degenerate coefficients attract Brownian motion. E = (x, φ(x)) R n. φ : R d R n d. David-Feneuil-Mayboroda: solutions to ( ) A(x) Lu = div u = 0, dist(x, E) n d 1 see the set E (A an elliptic matrix). 17

87 Degenerate Elliptic PDE Need degenerate elliptic PDE. Degenerate coefficients attract Brownian motion. E = (x, φ(x)) R n. φ : R d R n d. David-Feneuil-Mayboroda: solutions to ( ) A(x) Lu = div u = 0, dist(x, E) n d 1 see the set E (A an elliptic matrix) Red: Eau de Toilette: attracts the Brownian motion towards E. 17

88 Degenerate Elliptic PDE Need degenerate elliptic PDE. Degenerate coefficients attract Brownian motion. E = (x, φ(x)) R n. φ : R d R n d. David-Feneuil-Mayboroda: solutions to ( ) A(x) Lu = div u = 0, dist(x, E) n d 1 see the set E (A an elliptic matrix) Red: Eau de Toilette: attracts the Brownian motion towards E. Question: Geometry of E characterized by ω L vs σ? 17

89 Regularized Distance I: A Better Scent E = (x, φ(x)), φ(x) : R d R n d. Problem: x dist(x, E) is not a nice function. Hard to talk about ω L. 18

90 Regularized Distance I: A Better Scent E = (x, φ(x)), φ(x) : R d R n d. Problem: x dist(x, E) is not a nice function. Hard to talk about ω L. David-Feneuil-Mayboroda: family of smoothed out distances, D α (x). D α (x) dist(x, E). 18

91 Regularized Distance I: A Better Scent E = (x, φ(x)), φ(x) : R d R n d. Problem: x dist(x, E) is not a nice function. Hard to talk about ω L. David-Feneuil-Mayboroda: family of smoothed out distances, D α (x). D α (x) dist(x, E). Define (ˆ ) 1 1/α D α (x) dσ. x y d+α E 18

92 Regularized Distance I: A Better Scent E = (x, φ(x)), φ(x) : R d R n d. Problem: x dist(x, E) is not a nice function. Hard to talk about ω L. David-Feneuil-Mayboroda: family of smoothed out distances, D α (x). D α (x) dist(x, E). Define (ˆ ) 1 1/α D α (x) dσ. x y d+α α > 0 ensures D α (x) dist(x, E). E 18

93 Regularized Distance I: A Better Scent E = (x, φ(x)), φ(x) : R d R n d. Problem: x dist(x, E) is not a nice function. Hard to talk about ω L. David-Feneuil-Mayboroda: family of smoothed out distances, D α (x). D α (x) dist(x, E). Define (ˆ ) 1 1/α D α (x) dσ. x y d+α α > 0 ensures D α (x) dist(x, E). E D α sees whole geometry of E (non-local!) and is smooth in R n \E. 18

94 Regularized Distance I: A Better Scent E = (x, φ(x)), φ(x) : R d R n d. Problem: x dist(x, E) is not a nice function. Hard to talk about ω L. David-Feneuil-Mayboroda: family of smoothed out distances, D α (x). D α (x) dist(x, E). Define (ˆ ) 1 1/α D α (x) dσ. x y d+α α > 0 ensures D α (x) dist(x, E). E D α sees whole geometry of E (non-local!) and is smooth in R n \E. Oscillation of D α sees oscillation of φ (David-E.-Mayboroda 18) 18

95 Regularized Distance I: A Better Scent E = (x, φ(x)), φ(x) : R d R n d. Problem: x dist(x, E) is not a nice function. Hard to talk about ω L. David-Feneuil-Mayboroda: family of smoothed out distances, D α (x). D α (x) dist(x, E). Define (ˆ ) 1 1/α D α (x) dσ. x y d+α α > 0 ensures D α (x) dist(x, E). E D α sees whole geometry of E (non-local!) and is smooth in R n \E. Oscillation of D α sees oscillation of φ (David-E.-Mayboroda 18) Baby case! D α = constant iff φ 0. 18

96 Regularized Distance II: The Direct Result E = (x, φ(x)), φ : R d R n d. σ = surface measure and α > 0. Define (ˆ D α (x) E ) 1 1/α dσ(y). x y d+α ( ) 1 L α u div u. D α (x) n d 1 19

97 Regularized Distance II: The Direct Result E = (x, φ(x)), φ : R d R n d. σ = surface measure and α > 0. Define (ˆ D α (x) E ) 1 1/α dσ(y). x y d+α ( ) 1 L α u div u. D α (x) n d 1 Theorem (David-Feneuil-Mayboroda 2017) Let E be the graph of a Lipschitz φ : R d R n d with small Lip constant. Then ω X L α = k X dσ and k X is not too small or too big too often (A weight). 19

98 Regularized Distance II: The Direct Result E = (x, φ(x)), φ : R d R n d. σ = surface measure and α > 0. Define (ˆ D α (x) E ) 1 1/α dσ(y). x y d+α ( ) 1 L α u div u. D α (x) n d 1 Theorem (David-Feneuil-Mayboroda 2017) Let E be the graph of a Lipschitz φ : R d R n d with small Lip constant. Then ω X L α = k X dσ and k X is not too small or too big too often (A weight). Answers direct question in co-dimension > 1. 19

99 Regularized Distance II: The Direct Result E = (x, φ(x)), φ : R d R n d. σ = surface measure and α > 0. Define (ˆ D α (x) E ) 1 1/α dσ(y). x y d+α ( ) 1 L α u div u. D α (x) n d 1 Theorem (David-Feneuil-Mayboroda 2017) Let E be the graph of a Lipschitz φ : R d R n d with small Lip constant. Then ω X L α = k X dσ and k X is not too small or too big too often (A weight). Answers direct question in co-dimension > 1. Note: applies to much more general scents (i.e. any suitably smooth replacement for D α (x) (n d 1) I works). 19

100 What about the free boundary? E = (x, φ(x)) R n. If ω Lα = kdσ and k is nice does that mean that φ is nice? 20

101 What about the free boundary? E = (x, φ(x)) R n. If ω Lα = kdσ and k is nice does that mean that φ is nice? General Free Boundary Problem: Does the oscillation of k control the oscillation of φ? 20

102 What about the free boundary? E = (x, φ(x)) R n. If ω Lα = kdσ and k is nice does that mean that φ is nice? General Free Boundary Problem: Does the oscillation of k control the oscillation of φ? Baby Case: If k = constant must it be that φ = constant? 20

103 What about the free boundary? E = (x, φ(x)) R n. If ω Lα = kdσ and k is nice does that mean that φ is nice? General Free Boundary Problem: Does the oscillation of k control the oscillation of φ? Baby Case: If k = constant must it be that φ = constant? NO!!!! 20

104 Magic α! Let E = (x, φ(x)) R n, φ : R d R n d is Lipschitz. And α = n d 2 > 0. Theorem (David-E.-Mayboroda 18) For any E, α as above, have ω Lα = constant dσ. 21

105 Magic α! Let E = (x, φ(x)) R n, φ : R d R n d is Lipschitz. And α = n d 2 > 0. Theorem (David-E.-Mayboroda 18) For any E, α as above, have ω Lα = constant dσ. NOTE: A version for when E is fractal! d non-integer (here ω Lα σ). 21

106 Magic α! Let E = (x, φ(x)) R n, φ : R d R n d is Lipschitz. And α = n d 2 > 0. Theorem (David-E.-Mayboroda 18) For any E, α as above, have ω Lα = constant dσ. NOTE: A version for when E is fractal! d non-integer (here ω Lα σ). Recall in co-dimension 1: ω X = k X dσ, k X = constant Ω = B(X, R). 21

107 Magic α! Let E = (x, φ(x)) R n, φ : R d R n d is Lipschitz. And α = n d 2 > 0. Theorem (David-E.-Mayboroda 18) For any E, α as above, have ω Lα = constant dσ. NOTE: A version for when E is fractal! d non-integer (here ω Lα σ). Recall in co-dimension 1: ω X = k X dσ, k X = constant Ω = B(X, R). Takeaway: For magic α, dωα dσ doesn t control the regularity of φ, and fails to do so in the most spectacular way possible! 21

108 Magic α! Let E = (x, φ(x)) R n, φ : R d R n d is Lipschitz. And α = n d 2 > 0. Theorem (David-E.-Mayboroda 18) For any E, α as above, have ω Lα = constant dσ. NOTE: A version for when E is fractal! d non-integer (here ω Lα σ). Recall in co-dimension 1: ω X = k X dσ, k X = constant Ω = B(X, R). Takeaway: For magic α, dωα dσ doesn t control the regularity of φ, and fails to do so in the most spectacular way possible! D α is too nice a scent! 21

109 What s up with magic α? Can compute: see that for α = n d 2 we have ( ) 1 L α D α = div D α = 0. D n d 1 α The distance is a solution to the equation 22

110 What s up with magic α? Can compute: see that for α = n d 2 we have ( ) 1 L α D α = div D α = 0. D n d 1 α The distance is a solution to the equation D α is Green function with pole at infinity : D α on E gives dωα dσ. 22

111 What s up with magic α? Can compute: see that for α = n d 2 we have ( ) 1 L α D α = div D α = 0. D n d 1 α The distance is a solution to the equation D α is Green function with pole at infinity : D α on E gives dωα dσ. Note: In general computing the Green s function is VERY HARD! 22

112 What s up with magic α? Can compute: see that for α = n d 2 we have ( ) 1 L α D α = div D α = 0. D n d 1 α The distance is a solution to the equation D α is Green function with pole at infinity : D α on E gives dωα dσ. Note: In general computing the Green s function is VERY HARD! D α dist(x, E) ω α σ 22

113 What s up with magic α? Can compute: see that for α = n d 2 we have ( ) 1 L α D α = div D α = 0. D n d 1 α The distance is a solution to the equation D α is Green function with pole at infinity : D α on E gives dωα dσ. Note: In general computing the Green s function is VERY HARD! ( When α is magic D α (x) = D α dist(x, E) ω α σ E ) 1 1/α. dσ x y Note: 1 n 2 x n 2 is harmonic! 22

114 Open Questions about the Magic α 1 Why is magic α magic? D α satisfies an equation but what is really going on? 23

115 Open Questions about the Magic α 1 Why is magic α magic? D α satisfies an equation but what is really going on? Physical/geometric/probabilistic interpretation? 23

116 Open Questions about the Magic α 1 Why is magic α magic? D α satisfies an equation but what is really going on? Physical/geometric/probabilistic interpretation? 2 Is this emblematic or pathological? Is any other β magic? 23

117 Open Questions about the Magic α 1 Why is magic α magic? D α satisfies an equation but what is really going on? Physical/geometric/probabilistic interpretation? 2 Is this emblematic or pathological? Is any other β magic? Can we prove the converse for ω β with β not magic? 23

118 Open Questions about the Magic α 1 Why is magic α magic? D α satisfies an equation but what is really going on? Physical/geometric/probabilistic interpretation? 2 Is this emblematic or pathological? Is any other β magic? Can we prove the converse for ω β with β not magic? 3 What does α D α look like? The power 1 α makes this question harder. 23

119 Open Questions about the Magic α 1 Why is magic α magic? D α satisfies an equation but what is really going on? Physical/geometric/probabilistic interpretation? 2 Is this emblematic or pathological? Is any other β magic? Can we prove the converse for ω β with β not magic? 3 What does α D α look like? The power 1 α makes this question harder. 4 Can we do this in co-dimension one? Two? 23

120 Thanks! Thank You For Listening! The way of Laplace! 24