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Scattering by fractal screens: functional analysis and computation - PowerPoint PPT Presentation

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20 J UNE 2018, B OLOGNA Scattering by fractal screens: functional analysis and computation Andrea Moiola D IPARTIMENTO DI M ATEMATICA , U NIVERSIT DI P AVIA Joint work with S.N. Chandler-Wilde (Reading), D.P . Hewett (UCL) and A. Caetano

  1. Boundary integral equations (BIEs) BIEs provide a natural analytical and computational framework. x 3 (∆ + k 2 ) u = 0 , SRC x 2 Γ u = − u i or ∂ u /∂ n = − ∂ u i /∂ n x 1 loc ( R n + 1 \ Γ) ◮ Seek BVP solutions in W 1 ◮ Represent solutions in terms of jumps of boundary traces on Γ ◮ These jumps live in some ( Γ -dependent) subspaces of H ± 1 / 2 ( R n ) ◮ The jumps satisfy certain boundary integral equations ◮ The associated boundary integral operators are coercive, thus invertible, between appropriate spaces (Ha-Duong, Chandler-Wilde/Hewett) 9

  2. Boundary integral equations (BIEs) BIEs provide a natural analytical and computational framework. x 3 (∆ + k 2 ) u = 0 , SRC x 2 Γ u = − u i or ∂ u /∂ n = − ∂ u i /∂ n x 1 loc ( R n + 1 \ Γ) ◮ Seek BVP solutions in W 1 ◮ Represent solutions in terms of jumps of boundary traces on Γ ◮ These jumps live in some ( Γ -dependent) subspaces of H ± 1 / 2 ( R n ) ◮ The jumps satisfy certain boundary integral equations ◮ The associated boundary integral operators are coercive, thus invertible, between appropriate spaces (Ha-Duong, Chandler-Wilde/Hewett) 9

  3. Boundary integral equations (BIEs) BIEs provide a natural analytical and computational framework. x 3 (∆ + k 2 ) u = 0 , SRC x 2 Γ u = − u i or ∂ u /∂ n = − ∂ u i /∂ n x 1 loc ( R n + 1 \ Γ) ◮ Seek BVP solutions in W 1 ◮ Represent solutions in terms of jumps of boundary traces on Γ ◮ These jumps live in some ( Γ -dependent) subspaces of H ± 1 / 2 ( R n ) ◮ The jumps satisfy certain boundary integral equations ◮ The associated boundary integral operators are coercive, thus invertible, between appropriate spaces (Ha-Duong, Chandler-Wilde/Hewett) 9

  4. Sobolev spaces on Γ ⊂ R n BIEs require us to work in fractional (Bessel) Sobolev spaces on Γ ⊂ R n . For s ∈ R let � � � u ( ξ ) | 2 d ξ < ∞ u ∈ S ∗ ( R n ) : � u � 2 H s ( R n ) = R n ( 1 + | ξ | 2 ) s | ˆ H s ( R n ) := . For Γ ⊂ R n open and F ⊂ R n closed define [M C L EAN ] H s (Γ) := { u | Γ : u ∈ H s ( R n ) } restriction H s ( R n ) H s (Γ) := C ∞ � closure 0 (Γ) H s F := { u ∈ H s ( R n ) : supp u ⊂ F } support “Global” and “local” spaces: | Γ � H s (Γ) ⊂ H s ⊂ H s ( R n ) ⊂ D ∗ ( R n ) H s (Γ) ⊂ D ∗ (Γ) . − − − − − − − − − → Γ � �� � restriction oper. “0-trace” 10

  5. Sobolev spaces on Γ ⊂ R n BIEs require us to work in fractional (Bessel) Sobolev spaces on Γ ⊂ R n . For s ∈ R let � � � u ( ξ ) | 2 d ξ < ∞ u ∈ S ∗ ( R n ) : � u � 2 H s ( R n ) = R n ( 1 + | ξ | 2 ) s | ˆ H s ( R n ) := . For Γ ⊂ R n open and F ⊂ R n closed define [M C L EAN ] H s (Γ) := { u | Γ : u ∈ H s ( R n ) } restriction H s ( R n ) H s (Γ) := C ∞ � closure 0 (Γ) H s F := { u ∈ H s ( R n ) : supp u ⊂ F } support “Global” and “local” spaces: | Γ � H s (Γ) ⊂ H s ⊂ H s ( R n ) ⊂ D ∗ ( R n ) H s (Γ) ⊂ D ∗ (Γ) . − − − − − − − − − → Γ � �� � restriction oper. “0-trace” 10

  6. Sobolev spaces on Γ ⊂ R n BIEs require us to work in fractional (Bessel) Sobolev spaces on Γ ⊂ R n . For s ∈ R let � � � u ( ξ ) | 2 d ξ < ∞ u ∈ S ∗ ( R n ) : � u � 2 H s ( R n ) = R n ( 1 + | ξ | 2 ) s | ˆ H s ( R n ) := . For Γ ⊂ R n open and F ⊂ R n closed define [M C L EAN ] H s (Γ) := { u | Γ : u ∈ H s ( R n ) } restriction H s ( R n ) H s (Γ) := C ∞ � closure 0 (Γ) H s F := { u ∈ H s ( R n ) : supp u ⊂ F } support “Global” and “local” spaces: | Γ � H s (Γ) ⊂ H s ⊂ H s ( R n ) ⊂ D ∗ ( R n ) H s (Γ) ⊂ D ∗ (Γ) . − − − − − − − − − → Γ � �� � restriction oper. “0-trace” 10

  7. Properties of Sobolev spaces on Γ ⊂ R n When Γ is Lipschitz it holds that For general non-Lipschitz Γ H s (Γ) = ( H − s (Γ)) ∗ with equal norms ◮ � ◮ � � H s (Ω) ∼ � ◮ s ∈ N ⇒ � u � 2 Ω | ∂ α u | 2 ◮ × | α |≤ s ( ∼ ◮ � 00 (Γ) , s ≥ 0 ) H s (Γ) = H s = H s ◮ × Γ ◮ H ± 1 / 2 = { 0 } ◮ × ∂ Γ ◮ { H s (Γ) } s ∈ R and { � H s (Γ) } s ∈ R ◮ × are interpolation scales. This has implications for the scattering problem! There exist many works on Sobolev (Besov,. . . ) spaces on rough sets; most use intrinsic definitions on (e.g.) d -sets. Analogous to W s (Γ) , based on L p (Γ , H d ) . Related to spaces in R n by traces. See: Jonsson–Wallin, Strichartz. Our spaces are different, more suited for integral equations and BEM. 11

  8. Properties of Sobolev spaces on Γ ⊂ R n When Γ is Lipschitz it holds that For general non-Lipschitz Γ H s (Γ) = ( H − s (Γ)) ∗ with equal norms ◮ � ◮ � � H s (Ω) ∼ � ◮ s ∈ N ⇒ � u � 2 Ω | ∂ α u | 2 ◮ × | α |≤ s ( ∼ ◮ � 00 (Γ) , s ≥ 0 ) H s (Γ) = H s = H s ◮ × Γ ◮ H ± 1 / 2 = { 0 } ◮ × ∂ Γ ◮ { H s (Γ) } s ∈ R and { � H s (Γ) } s ∈ R ◮ × are interpolation scales. This has implications for the scattering problem! There exist many works on Sobolev (Besov,. . . ) spaces on rough sets; most use intrinsic definitions on (e.g.) d -sets. Analogous to W s (Γ) , based on L p (Γ , H d ) . Related to spaces in R n by traces. See: Jonsson–Wallin, Strichartz. Our spaces are different, more suited for integral equations and BEM. 11

  9. Properties of Sobolev spaces on Γ ⊂ R n When Γ is Lipschitz it holds that For general non-Lipschitz Γ H s (Γ) = ( H − s (Γ)) ∗ with equal norms ◮ � ◮ � � H s (Ω) ∼ � ◮ s ∈ N ⇒ � u � 2 Ω | ∂ α u | 2 ◮ × | α |≤ s ( ∼ ◮ � 00 (Γ) , s ≥ 0 ) H s (Γ) = H s = H s ◮ × Γ ◮ H ± 1 / 2 = { 0 } ◮ × ∂ Γ ◮ { H s (Γ) } s ∈ R and { � H s (Γ) } s ∈ R ◮ × are interpolation scales. This has implications for the scattering problem! There exist many works on Sobolev (Besov,. . . ) spaces on rough sets; most use intrinsic definitions on (e.g.) d -sets. Analogous to W s (Γ) , based on L p (Γ , H d ) . Related to spaces in R n by traces. See: Jonsson–Wallin, Strichartz. Our spaces are different, more suited for integral equations and BEM. 11

  10. Dirichlet BVP (Lipschitz open Γ ⊂ R n ) Problem D u i | Γ ), find u ∈ C 2 ( D ) ∩ W 1 Given g D ∈ H 1 / 2 (Γ) (e.g. g D = − loc ( D ) such that in D = R n + 1 \ Γ , D ⊂ R n + 1 (∆ + k 2 ) u = 0 on Γ , u = g D Γ ⊂ R n and u satisfies the Sommerfeld radiation condition. Theorem (cf. Stephan and Wendland ’84, Stephan ’87) If Γ is Lipschitz then D has a unique solution for all g D ∈ H 1 / 2 (Γ) . single-layer BIE: representation: S [ ∂ n u ] = − g D u = −S [ ∂ n u ] potential ( S ) operator ( S ): � S : � H − 1 / 2 (Γ) → C 2 ( D ) ∩ W 1 loc ( D ) S φ ( x ) := Φ( x , y ) φ ( y ) d s ( y ) , x ∈ D Γ S : � H − 1 / 2 (Γ) → H 1 / 2 (Γ) S φ ( x ) := γ ± S φ | Γ ( x ) x ∈ Γ S invertible, Φ( x , y ) := e i k | x − y | / 4 π | x − y | ( in 3D ) 12

  11. Dirichlet BVP (Lipschitz open Γ ⊂ R n ) Problem D u i | Γ ), find u ∈ C 2 ( D ) ∩ W 1 Given g D ∈ H 1 / 2 (Γ) (e.g. g D = − loc ( D ) such that in D = R n + 1 \ Γ , D ⊂ R n + 1 (∆ + k 2 ) u = 0 ( γ ± u ) | Γ = g D , Γ ⊂ R n and u satisfies the Sommerfeld radiation condition. Theorem (cf. Stephan and Wendland ’84, Stephan ’87) If Γ is Lipschitz then D has a unique solution for all g D ∈ H 1 / 2 (Γ) . single-layer BIE: S [ ∂ n u ] = − g D representation: u = −S [ ∂ n u ] potential ( S ) operator ( S ): � S : � H − 1 / 2 (Γ) → C 2 ( D ) ∩ W 1 loc ( D ) S φ ( x ) := Φ( x , y ) φ ( y ) d s ( y ) , x ∈ D Γ S : � H − 1 / 2 (Γ) → H 1 / 2 (Γ) S φ ( x ) := γ ± S φ | Γ ( x ) x ∈ Γ S invertible, Φ( x , y ) := e i k | x − y | / 4 π | x − y | ( in 3D ) 12

  12. Dirichlet BVP (Lipschitz open Γ ⊂ R n ) Problem D u i | Γ ), find u ∈ C 2 ( D ) ∩ W 1 Given g D ∈ H 1 / 2 (Γ) (e.g. g D = − loc ( D ) such that in D = R n + 1 \ Γ , D ⊂ R n + 1 (∆ + k 2 ) u = 0 ( γ ± u ) | Γ = g D , Γ ⊂ R n and u satisfies the Sommerfeld radiation condition. Theorem (cf. Stephan and Wendland ’84, Stephan ’87) If Γ is Lipschitz then D has a unique solution for all g D ∈ H 1 / 2 (Γ) . single-layer BIE: S [ ∂ n u ] = − g D representation: u = −S [ ∂ n u ] potential ( S ) operator ( S ): � S : � H − 1 / 2 (Γ) → C 2 ( D ) ∩ W 1 loc ( D ) S φ ( x ) := Φ( x , y ) φ ( y ) d s ( y ) , x ∈ D Γ S : � H − 1 / 2 (Γ) → H 1 / 2 (Γ) S φ ( x ) := γ ± S φ | Γ ( x ) x ∈ Γ S invertible, Φ( x , y ) := e i k | x − y | / 4 π | x − y | ( in 3D ) 12

  13. Dirichlet BVP (Lipschitz open Γ ⊂ R n ) Problem D u i | Γ ), find u ∈ C 2 ( D ) ∩ W 1 Given g D ∈ H 1 / 2 (Γ) (e.g. g D = − loc ( D ) such that in D = R n + 1 \ Γ , D ⊂ R n + 1 (∆ + k 2 ) u = 0 ( γ ± u ) | Γ = g D , Γ ⊂ R n and u satisfies the Sommerfeld radiation condition. Theorem (cf. Stephan and Wendland ’84, Stephan ’87) If Γ is Lipschitz then D has a unique solution for all g D ∈ H 1 / 2 (Γ) . single-layer BIE: S [ ∂ n u ] = − g D representation: u = −S [ ∂ n u ] potential ( S ) operator ( S ): � S : � H − 1 / 2 (Γ) → C 2 ( D ) ∩ W 1 loc ( D ) S φ ( x ) := Φ( x , y ) φ ( y ) d s ( y ) , x ∈ D Γ S : � H − 1 / 2 (Γ) → H 1 / 2 (Γ) S φ ( x ) := γ ± S φ | Γ ( x ) x ∈ Γ S invertible, Φ( x , y ) := e i k | x − y | / 4 π | x − y | ( in 3D ) 12

  14. Failure of BVP D for non-Lipschitz Γ What if Γ is not Lipschitz? Still have existence, but in general have non-uniqueness: [ ∂ n u ] ∈ H − 1 / 2 and [ u ] ∈ H 1 / 2 ◮ By Helmholtz eq.: . By BCs: Γ Γ [ u ] ∈ H 1 / 2 ∂ Γ ⊂ H 1 / 2 ( γ + u ) | Γ = g D = ( γ − u ) | Γ ⇒ [ u ] | Γ = 0 ⇒ . Γ If ∃ 0 � = φ ∈ H 1 / 2 ∂ Γ then D φ satisfies homogeneous problem. ( D = double layer potential.) H − 1 / 2 (Γ) � = H − 1 / 2 then ∃ 0 � = φ ∈ H − 1 / 2 ◮ If � \ � H − 1 / 2 (Γ) with S φ = 0 Γ Γ (S extended to S : H − 1 / 2 → H 1 / 2 (Γ) , continuous but not injective) Γ Then S φ satisfies homogeneous problem. We need to modify D to deal with this. 13

  15. Failure of BVP D for non-Lipschitz Γ What if Γ is not Lipschitz? Still have existence, but in general have non-uniqueness: [ ∂ n u ] ∈ H − 1 / 2 and [ u ] ∈ H 1 / 2 ◮ By Helmholtz eq.: . By BCs: Γ Γ [ u ] ∈ H 1 / 2 ∂ Γ ⊂ H 1 / 2 ( γ + u ) | Γ = g D = ( γ − u ) | Γ ⇒ [ u ] | Γ = 0 ⇒ . Γ If ∃ 0 � = φ ∈ H 1 / 2 ∂ Γ then D φ satisfies homogeneous problem. ( D = double layer potential.) H − 1 / 2 (Γ) � = H − 1 / 2 then ∃ 0 � = φ ∈ H − 1 / 2 ◮ If � \ � H − 1 / 2 (Γ) with S φ = 0 Γ Γ (S extended to S : H − 1 / 2 → H 1 / 2 (Γ) , continuous but not injective) Γ Then S φ satisfies homogeneous problem. We need to modify D to deal with this. 13

  16. Failure of BVP D for non-Lipschitz Γ What if Γ is not Lipschitz? Still have existence, but in general have non-uniqueness: [ ∂ n u ] ∈ H − 1 / 2 and [ u ] ∈ H 1 / 2 ◮ By Helmholtz eq.: . By BCs: Γ Γ [ u ] ∈ H 1 / 2 ∂ Γ ⊂ H 1 / 2 ( γ + u ) | Γ = g D = ( γ − u ) | Γ ⇒ [ u ] | Γ = 0 ⇒ . Γ If ∃ 0 � = φ ∈ H 1 / 2 ∂ Γ then D φ satisfies homogeneous problem. ( D = double layer potential.) H − 1 / 2 (Γ) � = H − 1 / 2 then ∃ 0 � = φ ∈ H − 1 / 2 ◮ If � \ � H − 1 / 2 (Γ) with S φ = 0 Γ Γ (S extended to S : H − 1 / 2 → H 1 / 2 (Γ) , continuous but not injective) Γ Then S φ satisfies homogeneous problem. We need to modify D to deal with this. 13

  17. Failure of BVP D for non-Lipschitz Γ What if Γ is not Lipschitz? Still have existence, but in general have non-uniqueness: [ ∂ n u ] ∈ H − 1 / 2 and [ u ] ∈ H 1 / 2 ◮ By Helmholtz eq.: . By BCs: Γ Γ [ u ] ∈ H 1 / 2 ∂ Γ ⊂ H 1 / 2 ( γ + u ) | Γ = g D = ( γ − u ) | Γ ⇒ [ u ] | Γ = 0 ⇒ . Γ If ∃ 0 � = φ ∈ H 1 / 2 ∂ Γ then D φ satisfies homogeneous problem. ( D = double layer potential.) H − 1 / 2 (Γ) � = H − 1 / 2 then ∃ 0 � = φ ∈ H − 1 / 2 ◮ If � \ � H − 1 / 2 (Γ) with S φ = 0 Γ Γ (S extended to S : H − 1 / 2 → H 1 / 2 (Γ) , continuous but not injective) Γ Then S φ satisfies homogeneous problem. We need to modify D to deal with this. 13

  18. Failure of BVP D for non-Lipschitz Γ What if Γ is not Lipschitz? Still have existence, but in general have non-uniqueness: [ ∂ n u ] ∈ H − 1 / 2 and [ u ] ∈ H 1 / 2 ◮ By Helmholtz eq.: . By BCs: Γ Γ [ u ] ∈ H 1 / 2 ∂ Γ ⊂ H 1 / 2 ( γ + u ) | Γ = g D = ( γ − u ) | Γ ⇒ [ u ] | Γ = 0 ⇒ . Γ If ∃ 0 � = φ ∈ H 1 / 2 ∂ Γ then D φ satisfies homogeneous problem. ( D = double layer potential.) H − 1 / 2 (Γ) � = H − 1 / 2 then ∃ 0 � = φ ∈ H − 1 / 2 ◮ If � \ � H − 1 / 2 (Γ) with S φ = 0 Γ Γ (S extended to S : H − 1 / 2 → H 1 / 2 (Γ) , continuous but not injective) Γ Then S φ satisfies homogeneous problem. We need to modify D to deal with this. 13

  19. Failure of BVP D for non-Lipschitz Γ What if Γ is not Lipschitz? Still have existence, but in general have non-uniqueness: [ ∂ n u ] ∈ H − 1 / 2 and [ u ] ∈ H 1 / 2 ◮ By Helmholtz eq.: . By BCs: Γ Γ [ u ] ∈ H 1 / 2 ∂ Γ ⊂ H 1 / 2 ( γ + u ) | Γ = g D = ( γ − u ) | Γ ⇒ [ u ] | Γ = 0 ⇒ . Γ If ∃ 0 � = φ ∈ H 1 / 2 ∂ Γ then D φ satisfies homogeneous problem. ( D = double layer potential.) H − 1 / 2 (Γ) � = H − 1 / 2 then ∃ 0 � = φ ∈ H − 1 / 2 ◮ If � \ � H − 1 / 2 (Γ) with S φ = 0 Γ Γ (S extended to S : H − 1 / 2 → H 1 / 2 (Γ) , continuous but not injective) Γ Then S φ satisfies homogeneous problem. We need to modify D to deal with this. 13

  20. Dirichlet BVP (arbitrary open Γ ) Problem � D u i | Γ ), find u ∈ C 2 ( D ) ∩ W 1 Given g D ∈ H 1 / 2 (Γ) (e.g. g D = − loc ( D ) such that (∆ + k 2 ) u = 0 in D , ( γ ± u ) | Γ = g D , ( D ′ ) [ u ] = 0 , � ∂ u � ∈ � ( D ′′ ) H − 1 / 2 (Γ) , ∂ n and u satisfies the Sommerfeld radiation condition. Theorem (Chandler-Wilde & Hewett 2013) For any bounded open Γ , � D has a unique solution for all g D ∈ H 1 / 2 (Γ) . then D ′ is superfluous. If H 1 / 2 ∂ Γ = { 0 } then D ′′ is superfluous. If � H − 1 / 2 (Γ) = H − 1 / 2 (E.g. if Γ is C 0 .) Γ Two key questions: (i) when is H s ∂ Γ = { 0 } ? (ii) when is � Γ ? H s (Γ) = H s 14

  21. Dirichlet BVP (arbitrary open Γ ) Problem � D u i | Γ ), find u ∈ C 2 ( D ) ∩ W 1 Given g D ∈ H 1 / 2 (Γ) (e.g. g D = − loc ( D ) such that (∆ + k 2 ) u = 0 in D , ( γ ± u ) | Γ = g D , ( D ′ ) [ u ] = 0 , � ∂ u � ∈ � ( D ′′ ) H − 1 / 2 (Γ) , ∂ n and u satisfies the Sommerfeld radiation condition. Theorem (Chandler-Wilde & Hewett 2013) For any bounded open Γ , � D has a unique solution for all g D ∈ H 1 / 2 (Γ) . then D ′ is superfluous. If H 1 / 2 ∂ Γ = { 0 } then D ′′ is superfluous. If � H − 1 / 2 (Γ) = H − 1 / 2 (E.g. if Γ is C 0 .) Γ Two key questions: (i) when is H s ∂ Γ = { 0 } ? (ii) when is � Γ ? H s (Γ) = H s 14

  22. Dirichlet BVP (arbitrary open Γ ) Problem � D u i | Γ ), find u ∈ C 2 ( D ) ∩ W 1 Given g D ∈ H 1 / 2 (Γ) (e.g. g D = − loc ( D ) such that (∆ + k 2 ) u = 0 in D , ( γ ± u ) | Γ = g D , ( D ′ ) [ u ] = 0 , � ∂ u � ∈ � ( D ′′ ) H − 1 / 2 (Γ) , ∂ n and u satisfies the Sommerfeld radiation condition. Theorem (Chandler-Wilde & Hewett 2013) For any bounded open Γ , � D has a unique solution for all g D ∈ H 1 / 2 (Γ) . then D ′ is superfluous. If H 1 / 2 ∂ Γ = { 0 } then D ′′ is superfluous. If � H − 1 / 2 (Γ) = H − 1 / 2 (E.g. if Γ is C 0 .) Γ Two key questions: (i) when is H s ∂ Γ = { 0 } ? (ii) when is � Γ ? H s (Γ) = H s 14

  23. Dirichlet BVP (arbitrary open Γ ) Problem � D u i | Γ ), find u ∈ C 2 ( D ) ∩ W 1 Given g D ∈ H 1 / 2 (Γ) (e.g. g D = − loc ( D ) such that (∆ + k 2 ) u = 0 in D , ( γ ± u ) | Γ = g D , ( D ′ ) [ u ] = 0 , � ∂ u � ∈ � ( D ′′ ) H − 1 / 2 (Γ) , ∂ n and u satisfies the Sommerfeld radiation condition. Theorem (Chandler-Wilde & Hewett 2013) For any bounded open Γ , � D has a unique solution for all g D ∈ H 1 / 2 (Γ) . then D ′ is superfluous. If H 1 / 2 ∂ Γ = { 0 } then D ′′ is superfluous. If � H − 1 / 2 (Γ) = H − 1 / 2 (E.g. if Γ is C 0 .) Γ Two key questions: (i) when is H s ∂ Γ = { 0 } ? (ii) when is � Γ ? H s (Γ) = H s 14

  24. Part II Two Sobolev space questions

  25. Key question #1: nullity Given a compact set K ⊂ R n with empty interior (e.g. K = ∂ Γ ), for which s ∈ R is H s K � = { 0 } ? Γ ∂ Γ Terminology: ⇒ ∄ non-zero elements of H s supported inside K . H s K = { 0 } ⇐ We call such a set K “ s -null”. Other terminology exists: “ ( − s ) -polar” (Maz’ya, Littman), “set of uniqueness for H s ” (Maz’ya, Adams/Hedberg). 15

  26. Key question #1: nullity Given a compact set K ⊂ R n with empty interior (e.g. K = ∂ Γ ), for which s ∈ R is H s K � = { 0 } ? Γ ∂ Γ Terminology: ⇒ ∄ non-zero elements of H s supported inside K . H s K = { 0 } ⇐ We call such a set K “ s -null”. Other terminology exists: “ ( − s ) -polar” (Maz’ya, Littman), “set of uniqueness for H s ” (Maz’ya, Adams/Hedberg). 15

  27. Nullity threshold For every compact K ⊂ R n with int ( K ) = ∅ , ∃ s K ∈ [ − n / 2 , n / 2 ] , called the nullity threshold of K , such that H s K = { 0 } for s > s K and H s K � = { 0 } for s < s K . H s H s K � = { 0 } K = { 0 } i.e. K supports H s distributions i.e. K cannot support H s distr. − n / 2 s K n / 2 s 0 Theorem (H & M 2017) Theorem (Polking 1972) If m ( K ) = 0 then ∃ compact K with int ( K ) = ∅ and m ( K ) > 0 for which s K = dim H ( K ) − n ≤ 0 H n / 2 � = { 0 } , so that s K = n / 2 . 2 K Connection with dim H comes from standard potential theory results (Maz’ya 2011, Adams & Hedberg 1996 etc.) Nullity theory ∼ complete for m ( K ) = 0 , open problems for m ( K ) > 0 . 16

  28. Nullity threshold For every compact K ⊂ R n with int ( K ) = ∅ , ∃ s K ∈ [ − n / 2 , n / 2 ] , called the nullity threshold of K , such that H s K = { 0 } for s > s K and H s K � = { 0 } for s < s K . H s H s K � = { 0 } K = { 0 } i.e. K supports H s distributions i.e. K cannot support H s distr. − n / 2 s K n / 2 s 0 Theorem (H & M 2017) Theorem (Polking 1972) If m ( K ) = 0 then ∃ compact K with int ( K ) = ∅ and m ( K ) > 0 for which s K = dim H ( K ) − n ≤ 0 H n / 2 � = { 0 } , so that s K = n / 2 . 2 K Connection with dim H comes from standard potential theory results (Maz’ya 2011, Adams & Hedberg 1996 etc.) Nullity theory ∼ complete for m ( K ) = 0 , open problems for m ( K ) > 0 . 16

  29. Key question #2: identity of 0-trace spaces Given an open set Γ ⊂ R n , when is � H s (Γ) = H s Γ ? Equivalent to density of C ∞ 0 (Γ) in { u ∈ H s ( R n ) : supp u ⊂ Γ } . Classical result (e.g. McLean) Let Γ ⊂ R n be C 0 . Then � H s (Γ) = H s Γ . 1st class of sets: “regular except at a few points”, e.g. prefractal Theorem (C-W, H & M 2017) Let n ≥ 2 , Γ ⊂ R n open and C 0 except at finite P ⊂ ∂ Γ . Then � H s (Γ) = H s Γ for | s | ≤ 1 . ◮ For n = 1 the same holds for | s | ≤ 1 / 2 . ◮ Can take countable P ⊂ ∂ Γ with finitely many limit points in every bounded subset of ∂ Γ . Proof uses sequence of special cutoffs for s = 1 , duality, interpolation. 17

  30. Key question #2: identity of 0-trace spaces Given an open set Γ ⊂ R n , when is � H s (Γ) = H s Γ ? Equivalent to density of C ∞ 0 (Γ) in { u ∈ H s ( R n ) : supp u ⊂ Γ } . Classical result (e.g. McLean) Let Γ ⊂ R n be C 0 . Then � H s (Γ) = H s Γ . 1st class of sets: “regular except at a few points”, e.g. prefractal Theorem (C-W, H & M 2017) Let n ≥ 2 , Γ ⊂ R n open and C 0 except at finite P ⊂ ∂ Γ . Then � H s (Γ) = H s Γ for | s | ≤ 1 . ◮ For n = 1 the same holds for | s | ≤ 1 / 2 . ◮ Can take countable P ⊂ ∂ Γ with finitely many limit points in every bounded subset of ∂ Γ . Proof uses sequence of special cutoffs for s = 1 , duality, interpolation. 17

  31. Key question #2: identity of 0-trace spaces Given an open set Γ ⊂ R n , when is � H s (Γ) = H s Γ ? Equivalent to density of C ∞ 0 (Γ) in { u ∈ H s ( R n ) : supp u ⊂ Γ } . Classical result (e.g. McLean) Let Γ ⊂ R n be C 0 . Then � H s (Γ) = H s Γ . 1st class of sets: “regular except at a few points”, e.g. prefractal Theorem (C-W, H & M 2017) Let n ≥ 2 , Γ ⊂ R n open and C 0 except at finite P ⊂ ∂ Γ . Then � H s (Γ) = H s Γ for | s | ≤ 1 . ◮ For n = 1 the same holds for | s | ≤ 1 / 2 . ◮ Can take countable P ⊂ ∂ Γ with finitely many limit points in every bounded subset of ∂ Γ . Proof uses sequence of special cutoffs for s = 1 , duality, interpolation. 17

  32. Examples of non- C 0 sets with � Γ , | s | ≤ 1 H s (Γ) = H s E.g. union of disjoint C 0 open sets, whose closures intersect only in P . Sierpinski triangle prefractals, (unbounded) checkerboard, double brick, inner and outer (double) curved cusps, spiral, Fraenkel’s “rooms and passages”. 18

  33. Constructing counterexamples Consider another class of sets: “nice domain minus small holes”. E.g. when int (Γ) is smooth. Theorem (C-W, H & M 2017) If int (Γ) is C 0 then � H s (Γ) = H s Γ ⇐ ⇒ int (Γ) \ Γ is ( − s ) -null. Corollary For every n ∈ N , there exists a bounded open set Γ ⊂ R n such that, H s (Γ) � H s � Γ , ∀ s ≥ − n / 2 Proof: take a ball and remove a Polking set (not s -null for any s ≤ n / 2 ) H s (Γ) � { u ∈ H s : u = 0 a.e. in Γ c } � H s � (Can also have ∀ s > 0 .) Γ 19

  34. Constructing counterexamples Consider another class of sets: “nice domain minus small holes”. E.g. when int (Γ) is smooth. Theorem (C-W, H & M 2017) If int (Γ) is C 0 then � H s (Γ) = H s Γ ⇐ ⇒ int (Γ) \ Γ is ( − s ) -null. Corollary For every n ∈ N , there exists a bounded open set Γ ⊂ R n such that, H s (Γ) � H s � Γ , ∀ s ≥ − n / 2 Proof: take a ball and remove a Polking set (not s -null for any s ≤ n / 2 ) H s (Γ) � { u ∈ H s : u = 0 a.e. in Γ c } � H s � (Can also have ∀ s > 0 .) Γ 19

  35. Constructing counterexamples Consider another class of sets: “nice domain minus small holes”. E.g. when int (Γ) is smooth. Theorem (C-W, H & M 2017) If int (Γ) is C 0 then � H s (Γ) = H s Γ ⇐ ⇒ int (Γ) \ Γ is ( − s ) -null. Corollary For every n ∈ N , there exists a bounded open set Γ ⊂ R n such that, H s (Γ) � H s � Γ , ∀ s ≥ − n / 2 Proof: take a ball and remove a Polking set (not s -null for any s ≤ n / 2 ) H s (Γ) � { u ∈ H s : u = 0 a.e. in Γ c } � H s � (Can also have ∀ s > 0 .) Γ 19

  36. Part III Formulations on general screens

  37. Prefractal convergence Theorem (C-W, H & M 2017) Consider a bounded sequence of nested open screens Γ 1 ⊂ Γ 2 ⊂ · · · For each j let u j denote the solution of problem � D for Γ j . Let Γ := � j ∈ N Γ j and let u denote the solution of problem � D for Γ . Then u j → u as j → ∞ (in W 1 loc ( D ) ). H s � � � Proof: � H s (Γ 1 ) ⊂ � � � � H s (Γ 2 ) ⊂ · · · and Γ j = H s (Γ j ) . j ∈ N j ∈ N Then write BIEs in variational form and apply Céa’s Lemma. What if we want to use Γ 1 ⊃ Γ 2 ⊃ · · · → Γ ? e.g. Cantor dust Need framework for closed screens. 20

  38. Prefractal convergence Theorem (C-W, H & M 2017) Consider a bounded sequence of nested open screens Γ 1 ⊂ Γ 2 ⊂ · · · For each j let u j denote the solution of problem � D for Γ j . Let Γ := � j ∈ N Γ j and let u denote the solution of problem � D for Γ . Then u j → u as j → ∞ (in W 1 loc ( D ) ). H s � � � Proof: � H s (Γ 1 ) ⊂ � � � � H s (Γ 2 ) ⊂ · · · and Γ j = H s (Γ j ) . j ∈ N j ∈ N Then write BIEs in variational form and apply Céa’s Lemma. What if we want to use Γ 1 ⊃ Γ 2 ⊃ · · · → Γ ? e.g. Cantor dust Need framework for closed screens. 20

  39. Prefractal convergence Theorem (C-W, H & M 2017) Consider a bounded sequence of nested open screens Γ 1 ⊂ Γ 2 ⊂ · · · For each j let u j denote the solution of problem � D for Γ j . Let Γ := � j ∈ N Γ j and let u denote the solution of problem � D for Γ . Then u j → u as j → ∞ (in W 1 loc ( D ) ). H s � � � Proof: � H s (Γ 1 ) ⊂ � � � � H s (Γ 2 ) ⊂ · · · and Γ j = H s (Γ j ) . j ∈ N j ∈ N Then write BIEs in variational form and apply Céa’s Lemma. What if we want to use Γ 1 ⊃ Γ 2 ⊃ · · · → Γ ? e.g. Cantor dust Need framework for closed screens. 20

  40. Prefractal convergence Theorem (C-W, H & M 2017) Consider a bounded sequence of nested open screens Γ 1 ⊂ Γ 2 ⊂ · · · For each j let u j denote the solution of problem � D for Γ j . Let Γ := � j ∈ N Γ j and let u denote the solution of problem � D for Γ . Then u j → u as j → ∞ (in W 1 loc ( D ) ). H s � � � Proof: � H s (Γ 1 ) ⊂ � � � � H s (Γ 2 ) ⊂ · · · and Γ j = H s (Γ j ) . j ∈ N j ∈ N Then write BIEs in variational form and apply Céa’s Lemma. What if we want to use Γ 1 ⊃ Γ 2 ⊃ · · · → Γ ? e.g. Cantor dust Need framework for closed screens. 20

  41. What about general screens? For an open screen Γ , we imposed the BC by restriction to Γ : ( γ ± u ) | Γ = g D H − 1 / 2 (Γ) → H 1 / 2 (Γ) ∼ S : � = ( � and viewed S as an operator H − 1 / 2 (Γ)) ∗ . | Γ ( H 1 / 2 H 1 / 2 ( R n ) Γ c ) ⊥ H 1 / 2 (Γ) But since ⊃ − − − − − − − → isomorphism we could equivalently impose the BC by orthogonal projection: Γ c ) ⊥ ( γ ± u ) = g D P ( H 1 / 2 Γ c ) ⊥ ∼ H − 1 / 2 (Γ) → ( H 1 / 2 and view S as an operator S : � = ( � H − 1 / 2 (Γ)) ∗ . This viewpoint suggests a way of writing down BVP formulations for general screens (even with int (Γ) = ∅ ): H − 1 / 2 (Γ) by some V − ⊂ H − 1 / 2 ( R n ) ◮ replace � ◮ characterise ( V − ) ∗ as a subspace V + ∗ ⊂ H 1 / 2 ( R n ) ◮ impose BC by orthogonal projection onto V + ∗ ◮ view S as an operator S : V − → V + ∗ 21

  42. What about general screens? For an open screen Γ , we imposed the BC by restriction to Γ : ( γ ± u ) | Γ = g D H − 1 / 2 (Γ) → H 1 / 2 (Γ) ∼ S : � = ( � and viewed S as an operator H − 1 / 2 (Γ)) ∗ . | Γ ( H 1 / 2 H 1 / 2 ( R n ) Γ c ) ⊥ H 1 / 2 (Γ) But since ⊃ − − − − − − − → isomorphism we could equivalently impose the BC by orthogonal projection: Γ c ) ⊥ ( γ ± u ) = g D P ( H 1 / 2 Γ c ) ⊥ ∼ H − 1 / 2 (Γ) → ( H 1 / 2 and view S as an operator S : � = ( � H − 1 / 2 (Γ)) ∗ . This viewpoint suggests a way of writing down BVP formulations for general screens (even with int (Γ) = ∅ ): H − 1 / 2 (Γ) by some V − ⊂ H − 1 / 2 ( R n ) ◮ replace � ◮ characterise ( V − ) ∗ as a subspace V + ∗ ⊂ H 1 / 2 ( R n ) ◮ impose BC by orthogonal projection onto V + ∗ ◮ view S as an operator S : V − → V + ∗ 21

  43. What about general screens? For an open screen Γ , we imposed the BC by restriction to Γ : ( γ ± u ) | Γ = g D H − 1 / 2 (Γ) → H 1 / 2 (Γ) ∼ S : � = ( � and viewed S as an operator H − 1 / 2 (Γ)) ∗ . | Γ ( H 1 / 2 H 1 / 2 ( R n ) Γ c ) ⊥ H 1 / 2 (Γ) But since ⊃ − − − − − − − → isomorphism we could equivalently impose the BC by orthogonal projection: Γ c ) ⊥ ( γ ± u ) = g D P ( H 1 / 2 Γ c ) ⊥ ∼ H − 1 / 2 (Γ) → ( H 1 / 2 and view S as an operator S : � = ( � H − 1 / 2 (Γ)) ∗ . This viewpoint suggests a way of writing down BVP formulations for general screens (even with int (Γ) = ∅ ): H − 1 / 2 (Γ) by some V − ⊂ H − 1 / 2 ( R n ) ◮ replace � ◮ characterise ( V − ) ∗ as a subspace V + ∗ ⊂ H 1 / 2 ( R n ) ◮ impose BC by orthogonal projection onto V + ∗ ◮ view S as an operator S : V − → V + ∗ 21

  44. Dirichlet BVP for general screens Let Γ be an arbitrary bounded subset of R n (not necessarily open). Let V − be any closed subspace of H − 1 / 2 ( R n ) satisfying H − 1 / 2 ( int (Γ)) ⊂ V − ⊂ H − 1 / 2 � , Γ = ( V − ) ∗ by V + ∗ := (( V − ) a ) ⊥ ⊂ H 1 / 2 ( R n ) . ∗ ∼ and define V + 22

  45. Dirichlet BVP for general screens Let Γ be an arbitrary bounded subset of R n (not necessarily open). Let V − be any closed subspace of H − 1 / 2 ( R n ) satisfying H − 1 / 2 ( int (Γ)) ⊂ V − ⊂ H − 1 / 2 � , Γ = ( V − ) ∗ by V + ∗ := (( V − ) a ) ⊥ ⊂ H 1 / 2 ( R n ) . ∗ ∼ and define V + 22

  46. Dirichlet BVP for general screens Let Γ be an arbitrary bounded subset of R n (not necessarily open). Let V − be any closed subspace of H − 1 / 2 ( R n ) satisfying H − 1 / 2 ( int (Γ)) ⊂ V − ⊂ H − 1 / 2 � , Γ = ( V − ) ∗ by V + ∗ := (( V − ) a ) ⊥ ⊂ H 1 / 2 ( R n ) . ∗ ∼ and define V + Here we are using the following fact: Let H , H be Hilbert spaces with H ∗ ∼ = H (unit. isom.). (E.g. H = H − 1 / 2 ( R n ) , H = H 1 / 2 ( R n ) .) = ( V a , H ) ⊥ , H (with inherited duality If V ⊂ H is a closed subspace, V ∗ ∼ pairing) 22

  47. Dirichlet BVP for general screens Let Γ be an arbitrary bounded subset of R n (not necessarily open). Let V − be any closed subspace of H − 1 / 2 ( R n ) satisfying H − 1 / 2 ( int (Γ)) ⊂ V − ⊂ H − 1 / 2 � , Γ = ( V − ) ∗ by V + ∗ := (( V − ) a ) ⊥ ⊂ H 1 / 2 ( R n ) . ∗ ∼ and define V + Problem D ( V − ) Given g D ∈ V + ∗ (e.g. g D = − P V + ∗ u i ), find u ∈ C 2 ( D ) ∩ W 1 loc ( D ) such that (∆ + k 2 ) u = 0 in D , ∗ γ ± u = g D , P V + [ u ] = 0 , [ ∂ n u ] ∈ V − , SRC at infinity. 22

  48. Dirichlet BVP for general screens Let Γ be an arbitrary bounded subset of R n (not necessarily open). Let V − be any closed subspace of H − 1 / 2 ( R n ) satisfying H − 1 / 2 ( int (Γ)) ⊂ V − ⊂ H − 1 / 2 � , Γ = ( V − ) ∗ by V + ∗ := (( V − ) a ) ⊥ ⊂ H 1 / 2 ( R n ) . ∗ ∼ and define V + Problem D ( V − ) Given g D ∈ V + ∗ (e.g. g D = − P V + ∗ u i ), Theorem (C-W & H 2016) find u ∈ C 2 ( D ) ∩ W 1 loc ( D ) such that Problem D ( V − ) is well-posed (∆ + k 2 ) u = 0 in D , for any choice of V − . ∗ γ ± u = g D , P V + S : V − → V + Operator [ u ] = 0 , ∗ inherits coercivity! [ ∂ n u ] ∈ V − , SRC at infinity. 22

  49. Which formulation to use? H − 1 / 2 ( int (Γ)) ⊂ V − ⊂ H − 1 / 2 � For any bounded Γ , each choice Γ gives its own well-posed formulation D ( V − ) . Theorem (C-W & H 2018) H − 1 / 2 ( int (Γ)) = H − 1 / 2 If � there is only one such formulation. Γ H − 1 / 2 ( int (Γ)) � = H − 1 / 2 If � ∃ infinitely many formulations with � = solutions! Γ To select “physically correct” solut., apply limiting geometry principle: • Γ 1 ⊂ Γ 2 ⊂ · · · open and “nice” • Γ 1 ⊃ Γ 2 ⊃ · · · closed and “nice” (e.g. Lipschitz) (e.g. closure of Lipschitz) • Γ := � • Γ := � j Γ j open (gray part), j Γ j closed (black part), → natural choice is → natural choice is V − = � V − = H − 1 / 2 H − 1 / 2 (Γ) . . Γ 23

  50. Which formulation to use? H − 1 / 2 ( int (Γ)) ⊂ V − ⊂ H − 1 / 2 � For any bounded Γ , each choice Γ gives its own well-posed formulation D ( V − ) . Theorem (C-W & H 2018) H − 1 / 2 ( int (Γ)) = H − 1 / 2 If � there is only one such formulation. Γ H − 1 / 2 ( int (Γ)) � = H − 1 / 2 If � ∃ infinitely many formulations with � = solutions! Γ To select “physically correct” solut., apply limiting geometry principle: • Γ 1 ⊂ Γ 2 ⊂ · · · open and “nice” • Γ 1 ⊃ Γ 2 ⊃ · · · closed and “nice” (e.g. Lipschitz) (e.g. closure of Lipschitz) • Γ := � • Γ := � j Γ j open (gray part), j Γ j closed (black part), → natural choice is → natural choice is V − = � V − = H − 1 / 2 H − 1 / 2 (Γ) . . Γ 23

  51. Which formulation to use? H − 1 / 2 ( int (Γ)) ⊂ V − ⊂ H − 1 / 2 � For any bounded Γ , each choice Γ gives its own well-posed formulation D ( V − ) . Theorem (C-W & H 2018) H − 1 / 2 ( int (Γ)) = H − 1 / 2 If � there is only one such formulation. Γ H − 1 / 2 ( int (Γ)) � = H − 1 / 2 If � ∃ infinitely many formulations with � = solutions! Γ To select “physically correct” solut., apply limiting geometry principle: • Γ 1 ⊂ Γ 2 ⊂ · · · open and “nice” • Γ 1 ⊃ Γ 2 ⊃ · · · closed and “nice” (e.g. Lipschitz) (e.g. closure of Lipschitz) • Γ := � • Γ := � j Γ j open (gray part), j Γ j closed (black part), → natural choice is → natural choice is V − = � V − = H − 1 / 2 H − 1 / 2 (Γ) . . Γ 23

  52. Which formulation to use? H − 1 / 2 ( int (Γ)) ⊂ V − ⊂ H − 1 / 2 � For any bounded Γ , each choice Γ gives its own well-posed formulation D ( V − ) . Theorem (C-W & H 2018) H − 1 / 2 ( int (Γ)) = H − 1 / 2 If � there is only one such formulation. Γ H − 1 / 2 ( int (Γ)) � = H − 1 / 2 If � ∃ infinitely many formulations with � = solutions! Γ To select “physically correct” solut., apply limiting geometry principle: • Γ 1 ⊂ Γ 2 ⊂ · · · open and “nice” • Γ 1 ⊃ Γ 2 ⊃ · · · closed and “nice” (e.g. Lipschitz) (e.g. closure of Lipschitz) • Γ := � • Γ := � j Γ j open (gray part), j Γ j closed (black part), → natural choice is → natural choice is V − = � V − = H − 1 / 2 H − 1 / 2 (Γ) . . Γ 23

  53. Which formulation to use? H − 1 / 2 ( int (Γ)) ⊂ V − ⊂ H − 1 / 2 � For any bounded Γ , each choice Γ gives its own well-posed formulation D ( V − ) . Theorem (C-W & H 2018) H − 1 / 2 ( int (Γ)) = H − 1 / 2 If � there is only one such formulation. Γ H − 1 / 2 ( int (Γ)) � = H − 1 / 2 If � ∃ infinitely many formulations with � = solutions! Γ To select “physically correct” solut., apply limiting geometry principle: • Γ 1 ⊂ Γ 2 ⊂ · · · open and “nice” • Γ 1 ⊃ Γ 2 ⊃ · · · closed and “nice” (e.g. Lipschitz) (e.g. closure of Lipschitz) • Γ := � • Γ := � j Γ j open (gray part), j Γ j closed (black part), → natural choice is → natural choice is V − = � V − = H − 1 / 2 H − 1 / 2 (Γ) . . Γ 23

  54. What if prefractals are not nested? �⊂ What if prefractals Γ j are neither increasing nor decreasing? Γ j �⊃ Γ j + 1 Key tool is Mosco convergence (Mosco 1969): M V j , V closed subspaces of Hilbert space H , j ∈ N , then V j → V if: − − ◮ ∀ v ∈ V , j ∈ N , ∃ v j ∈ V j s.t. v j → v (strong approximability) ◮ ∀ ( j m ) subsequence of N , v j m ∈ V j m for m ∈ N , v j m ⇀ v , then v ∈ V (weak closure) H − 1 / 2 ( int (Γ)) ⊂ V ⊂ H − 1 / 2 Think: H = H − 1 / 2 ( R n ) , V j = � H − 1 / 2 (Γ j ) , � Γ Theorem (C-W, H & M, 2018) M → V ⊂ H − 1 / 2 ( R n ) then solution of D ( V j ) converges to sol.n of D ( V ) If V j − − H − 1 / 2 ( int (Γ)) = H − 1 / 2 Holds for square snowflake above with V = � Γ 24

  55. What if prefractals are not nested? �⊂ What if prefractals Γ j are neither increasing nor decreasing? Γ j �⊃ Γ j + 1 Key tool is Mosco convergence (Mosco 1969): M V j , V closed subspaces of Hilbert space H , j ∈ N , then V j → V if: − − ◮ ∀ v ∈ V , j ∈ N , ∃ v j ∈ V j s.t. v j → v (strong approximability) ◮ ∀ ( j m ) subsequence of N , v j m ∈ V j m for m ∈ N , v j m ⇀ v , then v ∈ V (weak closure) H − 1 / 2 ( int (Γ)) ⊂ V ⊂ H − 1 / 2 Think: H = H − 1 / 2 ( R n ) , V j = � H − 1 / 2 (Γ j ) , � Γ Theorem (C-W, H & M, 2018) M → V ⊂ H − 1 / 2 ( R n ) then solution of D ( V j ) converges to sol.n of D ( V ) If V j − − H − 1 / 2 ( int (Γ)) = H − 1 / 2 Holds for square snowflake above with V = � Γ 24

  56. What if prefractals are not nested? �⊂ What if prefractals Γ j are neither increasing nor decreasing? Γ j �⊃ Γ j + 1 Key tool is Mosco convergence (Mosco 1969): M V j , V closed subspaces of Hilbert space H , j ∈ N , then V j → V if: − − ◮ ∀ v ∈ V , j ∈ N , ∃ v j ∈ V j s.t. v j → v (strong approximability) ◮ ∀ ( j m ) subsequence of N , v j m ∈ V j m for m ∈ N , v j m ⇀ v , then v ∈ V (weak closure) H − 1 / 2 ( int (Γ)) ⊂ V ⊂ H − 1 / 2 Think: H = H − 1 / 2 ( R n ) , V j = � H − 1 / 2 (Γ j ) , � Γ Theorem (C-W, H & M, 2018) M → V ⊂ H − 1 / 2 ( R n ) then solution of D ( V j ) converges to sol.n of D ( V ) If V j − − H − 1 / 2 ( int (Γ)) = H − 1 / 2 Holds for square snowflake above with V = � Γ 24

  57. When is u = 0 ? Theorem (C-W & H 2018) Let Γ be closed with empty interior and let V − = H − 1 / 2 . Γ ◮ If dim H Γ < n − 1 then u = 0 for every incident direction d . ◮ If dim H Γ > n − 1 then u � = 0 for a.e. incident direction d . So both the Sierpinski triangle ( dim H = log 3 / log 2 ) and pentaflake √ ( dim H = log 6 / log (( 3 + 5 ) / 2 ) ) generate a non-zero scattered field: 25

  58. When is u = 0 ? Theorem (C-W & H 2018) Let Γ be closed with empty interior and let V − = H − 1 / 2 . Γ ◮ If dim H Γ < n − 1 then u = 0 for every incident direction d . ◮ If dim H Γ > n − 1 then u � = 0 for a.e. incident direction d . So both the Sierpinski triangle ( dim H = log 3 / log 2 ) and pentaflake √ ( dim H = log 6 / log (( 3 + 5 ) / 2 ) ) generate a non-zero scattered field: 25

  59. Back to the Cantor dust α := C α × C α ⊂ R 2 denote the “Cantor dust” ( 0 < α < 1 / 2 ): Let C 2 1 α Question : Is the scattered field u zero or non-zero for the 3D Dirich- let scattering problem with Γ = C 2 α ? log ( 4 ) dim H ( C 2 α ) = log ( 1 /α ) Answer : u = 0 , if 0 < α ≤ 1 / 4 ; u � = 0 , in general, if 1 / 4 < α < 1 / 2 . ( u = 0 for all α for Neumann BCs) 26

  60. Back to the Cantor dust α := C α × C α ⊂ R 2 denote the “Cantor dust” ( 0 < α < 1 / 2 ): Let C 2 1 α Question : Is the scattered field u zero or non-zero for the 3D Dirich- let scattering problem with Γ = C 2 α ? log ( 4 ) dim H ( C 2 α ) = log ( 1 /α ) Answer : u = 0 , if 0 < α ≤ 1 / 4 ; u � = 0 , in general, if 1 / 4 < α < 1 / 2 . ( u = 0 for all α for Neumann BCs) 26

  61. Part IV Numerical approximation

  62. Boundary element method (BEM) For each prefractal Γ j , the BIE S [ ∂ u /∂ n ] = − g D can be solved using a standard BEM space, e.g. piecewise constants on a mesh of width h j . Let w j denote the Galerkin BEM solution on Γ j . Let l j = α j be the width of each component of Γ j ( 4 j of them). Under certain assumptions on h j , we prove BEM convergence � u − w j � H − 1 / 2 ( R n ) → 0 . Follows from Mosco convergence of BEM spaces. This requires approximability ( ∀ v ∈ H − 1 / 2 ∃ v j ∈ � H − 1 / 2 (Γ j ) , v j → v ): Γ proved with mollification, L 2 projection, partition of unity, . . . 27

  63. Boundary element method (BEM) For each prefractal Γ j , the BIE S [ ∂ u /∂ n ] = − g D can be solved using a standard BEM space, e.g. piecewise constants on a mesh of width h j . Let w j denote the Galerkin BEM solution on Γ j . Let l j = α j be the width of each component of Γ j ( 4 j of them). Under certain assumptions on h j , we prove BEM convergence � u − w j � H − 1 / 2 ( R n ) → 0 . Follows from Mosco convergence of BEM spaces. This requires approximability ( ∀ v ∈ H − 1 / 2 ∃ v j ∈ � H − 1 / 2 (Γ j ) , v j → v ): Γ proved with mollification, L 2 projection, partition of unity, . . . 27

  64. Boundary element method (BEM) For each prefractal Γ j , the BIE S [ ∂ u /∂ n ] = − g D can be solved using a standard BEM space, e.g. piecewise constants on a mesh of width h j . Let w j denote the Galerkin BEM solution on Γ j . Let l j = α j be the width of each component of Γ j ( 4 j of them). Under certain assumptions on h j , we prove BEM convergence � u − w j � H − 1 / 2 ( R n ) → 0 . Follows from Mosco convergence of BEM spaces. This requires approximability ( ∀ v ∈ H − 1 / 2 ∃ v j ∈ � H − 1 / 2 (Γ j ) , v j → v ): Γ proved with mollification, L 2 projection, partition of unity, . . . 27

  65. Convergence results for the Cantor dust Theorem (C-W, H & M 2018) Γ is dense in H − 1 / 2 Suppose ∃ − 1 / 2 < t < 0 such that H t . Γ Then ∃ µ = µ ( t ) > 0 such that if h j / l j = O ( e − µ j ) then w j → u as j → ∞ . Certainly not sharp! ◮ h j / l j = O ( e − µ j ) is a severe restriction Γ ⊂ H − 1 / 2 ◮ Density assumption H t for some t > − 1 / 2 not yet verified Γ We can do better if we replace Γ j by “fattened” versions: ˜ Γ j = { x : dist ( x , Γ j ) < ε l j } for some 0 < ε < min { α, 1 2 − α } . Theorem (C-W, H & M 2018) If h j = o ( l j ) then ˜ w j → u as j → ∞ . Γ is dense in H − 1 / 2 We require condition weaker than h j = o ( l j ) if H t . Γ For simplicity, I’ll show results on prefractals for #DOF fixed but large. 28

  66. Convergence results for the Cantor dust Theorem (C-W, H & M 2018) Γ is dense in H − 1 / 2 Suppose ∃ − 1 / 2 < t < 0 such that H t . Γ Then ∃ µ = µ ( t ) > 0 such that if h j / l j = O ( e − µ j ) then w j → u as j → ∞ . Certainly not sharp! ◮ h j / l j = O ( e − µ j ) is a severe restriction Γ ⊂ H − 1 / 2 ◮ Density assumption H t for some t > − 1 / 2 not yet verified Γ We can do better if we replace Γ j by “fattened” versions: ˜ Γ j = { x : dist ( x , Γ j ) < ε l j } for some 0 < ε < min { α, 1 2 − α } . Theorem (C-W, H & M 2018) If h j = o ( l j ) then ˜ w j → u as j → ∞ . Γ is dense in H − 1 / 2 We require condition weaker than h j = o ( l j ) if H t . Γ For simplicity, I’ll show results on prefractals for #DOF fixed but large. 28

  67. Convergence results for the Cantor dust Theorem (C-W, H & M 2018) Γ is dense in H − 1 / 2 Suppose ∃ − 1 / 2 < t < 0 such that H t . Γ Then ∃ µ = µ ( t ) > 0 such that if h j / l j = O ( e − µ j ) then w j → u as j → ∞ . Certainly not sharp! ◮ h j / l j = O ( e − µ j ) is a severe restriction Γ ⊂ H − 1 / 2 ◮ Density assumption H t for some t > − 1 / 2 not yet verified Γ We can do better if we replace Γ j by “fattened” versions: ˜ Γ j = { x : dist ( x , Γ j ) < ε l j } for some 0 < ε < min { α, 1 2 − α } . Theorem (C-W, H & M 2018) If h j = o ( l j ) then ˜ w j → u as j → ∞ . Γ is dense in H − 1 / 2 We require condition weaker than h j = o ( l j ) if H t . Γ For simplicity, I’ll show results on prefractals for #DOF fixed but large. 28

  68. Convergence results for the Cantor dust Theorem (C-W, H & M 2018) Γ is dense in H − 1 / 2 Suppose ∃ − 1 / 2 < t < 0 such that H t . Γ Then ∃ µ = µ ( t ) > 0 such that if h j / l j = O ( e − µ j ) then w j → u as j → ∞ . Certainly not sharp! ◮ h j / l j = O ( e − µ j ) is a severe restriction Γ ⊂ H − 1 / 2 ◮ Density assumption H t for some t > − 1 / 2 not yet verified Γ We can do better if we replace Γ j by “fattened” versions: ˜ Γ j = { x : dist ( x , Γ j ) < ε l j } for some 0 < ε < min { α, 1 2 − α } . Theorem (C-W, H & M 2018) If h j = o ( l j ) then ˜ w j → u as j → ∞ . Γ is dense in H − 1 / 2 We require condition weaker than h j = o ( l j ) if H t . Γ For simplicity, I’ll show results on prefractals for #DOF fixed but large. 28

  69. Convergence results for the Cantor dust Theorem (C-W, H & M 2018) Γ is dense in H − 1 / 2 Suppose ∃ − 1 / 2 < t < 0 such that H t . Γ Then ∃ µ = µ ( t ) > 0 such that if h j / l j = O ( e − µ j ) then w j → u as j → ∞ . Certainly not sharp! ◮ h j / l j = O ( e − µ j ) is a severe restriction Γ ⊂ H − 1 / 2 ◮ Density assumption H t for some t > − 1 / 2 not yet verified Γ We can do better if we replace Γ j by “fattened” versions: ˜ Γ j = { x : dist ( x , Γ j ) < ε l j } for some 0 < ε < min { α, 1 2 − α } . Theorem (C-W, H & M 2018) If h j = o ( l j ) then ˜ w j → u as j → ∞ . Γ is dense in H − 1 / 2 We require condition weaker than h j = o ( l j ) if H t . Γ For simplicity, I’ll show results on prefractals for #DOF fixed but large. 28

  70. Numerical results: Cantor dust α = 1 / 3 ( u � = 0 ) k = 25 , 4096 DOFs, prefractal level 1 29

  71. Numerical results: Cantor dust α = 1 / 3 ( u � = 0 ) k = 25 , 4096 DOFs, prefractal level 2 29

  72. Numerical results: Cantor dust α = 1 / 3 ( u � = 0 ) k = 25 , 4096 DOFs, prefractal level 3 29

  73. Numerical results: Cantor dust α = 1 / 3 ( u � = 0 ) k = 25 , 4096 DOFs, prefractal level 4 29

  74. Numerical results: Cantor dust α = 1 / 3 ( u � = 0 ) k = 25 , 4096 DOFs, prefractal level 5 29

  75. Numerical results: Cantor dust α = 1 / 3 ( u � = 0 ) k = 25 , 4096 DOFs, prefractal level 6 29

  76. Numerical results: Cantor dust α = 0 . 1 ( u = 0 ) k = 25 , 4096 DOFs, prefractal level 1 30

  77. Numerical results: Cantor dust α = 0 . 1 ( u = 0 ) k = 25 , 4096 DOFs, prefractal level 2 30

  78. Numerical results: Cantor dust α = 0 . 1 ( u = 0 ) k = 25 , 4096 DOFs, prefractal level 3 30

  79. Numerical results: Cantor dust α = 0 . 1 ( u = 0 ) k = 25 , 4096 DOFs, prefractal level 4 30

  80. Numerical results: Cantor dust α = 0 . 1 ( u = 0 ) k = 25 , 4096 DOFs, prefractal level 5 30

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