Computing the constant in Friedrichs inequality

Transkript

1 Computing the constant in Friedrichs inequality Tomáš Vejchodský Institute of Mathematics, Žitná 25, Praha 1 February 8, 212, SIGA 212, Prague

2 Motivation Classical formulation: u = f in Ω, u = on Ω Weak formulation: V = H 1(Ω) u V : a(u, v) = F(v) v V Error bound: u h V u u h y u h + C F f + div y y H(div, Ω) Notation: a(u, v) = ( u, v) F(v) = (f, v) (ϕ, ψ) = ϕψ dx Ω Energy norm: e 2 = a(e, e) = ( e, e) = e 2

3 Friedrichs inequality Standard: v C F v v H 1 (Ω)

4 Friedrichs inequality Standard: Generalization: v C F v v C F v v H 1 (Ω) v V

5 Friedrichs inequality Standard: Generalization: v C F v v C F v v H 1 (Ω) v V Variants: v 2 = u A = (A u, u)

6 Friedrichs inequality Standard: Generalization: v C F v v C F v v H 1 (Ω) v V Variants: v 2 = u A = (A u, u) v 2 = u 2 A + u 2 c = (A u, u) + (cu, u)

7 Friedrichs inequality Standard: Generalization: v C F v v C F v v H 1 (Ω) v V Variants: v 2 = u A = (A u, u) v 2 = u 2 A + u 2 c = (A u, u) + (cu, u) V = H 1 (Ω) Ω V = {v H 1 (Ω) : v = on Γ} Ω

8 Relation with eigenvalues Friedrichs inequality: v C F v v V C F = sup v V Laplace eigenvalue problem u i = λ i u i in Ω, u i = on Ω Theorem: C 2 F = 1 λ 1 where λ 1 = min i λ i. v v

9 Relation with eigenvalues Friedrichs inequality: v C F v v V C F = sup v V Laplace eigenvalue problem u i = λ i u i in Ω, u i = on Ω Theorem: C 2 F = 1 λ 1 where λ 1 = min i λ i. v v Proof: Weak formulation: u i V : ( u i, v) = λ i (u i, v) v V v 2 λ 1 = inf v V v 2 1 λ 1 = sup v V v 2 v 2

10 Rayleigh Ritz approximation of λ 1 Weak formulation: u i V : ( u i, v) = λ i (u i, v) v V Rayleigh Ritz method: V h V, dim V h < ui h V h : ( ui h, v h ) = λ h i (ui h, v h ) v h V h Theorem: λ 1 λ h 1

11 Rayleigh Ritz approximation of λ 1 Weak formulation: u i V : ( u i, v) = λ i (u i, v) v V Rayleigh Ritz method: V h V, dim V h < ui h V h : ( ui h, v h ) = λ h i (ui h, v h ) v h V h Theorem: λ 1 λ h 1 Proof: v 2 λ 1 = inf v V v 2 v 2 inf v h V h v 2 = λ h 1

12 Rayleigh Ritz approximation of λ 1 Weak formulation: u i V : ( u i, v) = λ i (u i, v) v V Rayleigh Ritz method: V h V, dim V h < ui h V h : ( ui h, v h ) = λ h i (ui h, v h ) v h V h Theorem: λ 1 λ h 1 Proof: v 2 λ 1 = inf v V v 2 v 2 inf v h V h v 2 = λ h 1 Corollary: C h F C F

13 Lower bound on λ 1 Method of a priori-a posteriori inequalities. Theorem (Kuttler and Sigillito, 1978): Let H be a separable Hilbert space. Let A : H H be a symmetric operator with dense domain D(A). Other technical assumptions on A. Let λ and u D(A) be arbitrary. Consider w D(A) such that Aw = Au λ u. Then min i λ i λ λ i w H. u H Ussage: H = L 2 (Ω), A = min λ i λ i λ i w w C F u u

14 Algorithm min i λ i λ λ i C F w u Theorem: w u q + C F λ u + div q q H(div, Ω).

15 Algorithm min i λ i λ λ i C F Theorem: w u q + C F λ u + div q ( ) u q λ u + div q + C F u u q H(div, Ω).

16 Algorithm min i λ i λ λ i C F Theorem: w u q + C F λ u + div q ( ) u q λ u + div q + C F u u q H(div, Ω). Compute Rayleigh Ritz approximations λ h 1 and uh 1

17 Algorithm min i λ i λ λ i C F Theorem: w u q + C F λ u + div q ( ) u q λ u + div q + C F u u q H(div, Ω). Compute Rayleigh Ritz approximations λ h 1 and uh 1 Set λ = λ h 1 and u = u h 1

18 Algorithm min i λ i λ h 1 λ i C F ( u h 1 q u h 1 Theorem: w u q + C F λ u + div q λ h 1 + C uh 1 + div q ) F u1 h q H(div, Ω). Compute Rayleigh Ritz approximations λ h 1 and uh 1 Set λ = λ h 1 and u = u h 1

19 Algorithm min i λ i λ h 1 λ i C F ( u h 1 q u h 1 Theorem: w u q + C F λ u + div q λ h 1 + C uh 1 + div q ) F u1 h q H(div, Ω). Compute Rayleigh Ritz approximations λ h 1 and uh 1 Set λ = λ h 1 and u = u h 1 Find approximate minimizer q h H(div, Ω)

20 Algorithm min i λ i λ h 1 λ i C F ( u h 1 q h u h 1 Theorem: w u q + C F λ u + div q λ h 1 + C uh 1 + div q ) h F u1 h q H(div, Ω). Compute Rayleigh Ritz approximations λ h 1 and uh 1 Set λ = λ h 1 and u = u h 1 Find approximate minimizer q h H(div, Ω)

21 Algorithm min i λ i λ h 1 λ i C F ( u h 1 q h u h 1 Theorem: w u q + C F λ u + div q λ h 1 + C uh 1 + div q ) h F u1 h q H(div, Ω). Compute Rayleigh Ritz approximations λ h 1 and uh 1 Set λ = λ h 1 and u = u1 h Find approximate minimizer q h H(div, Ω) u 1 h α = q h λ h 1 u1 h, β = uh 1 + div q h u1 h, C F = 1 λ1

22 Algorithm min i λ i λ h 1 λ i C F ( u h 1 q h u h 1 Theorem: w u q + C F λ u + div q λ h 1 + C uh 1 + div q ) h F u1 h q H(div, Ω). Compute Rayleigh Ritz approximations λ h 1 and uh 1 Set λ = λ h 1 and u = u1 h Find approximate minimizer q h H(div, Ω) u 1 h α = q h λ h 1 u1 h, β = uh 1 + div q h u1 h, C F = 1 λh 1 λ 1 1 ( α + 1 ) β λ 1 λ1 λ1 λ1

23 Algorithm min i λ i λ h 1 λ i C F ( u h 1 q h u h 1 Theorem: w u q + C F λ u + div q λ h 1 + C uh 1 + div q ) h F u1 h q H(div, Ω). Compute Rayleigh Ritz approximations λ h 1 and uh 1 Set λ = λ h 1 and u = u1 h Find approximate minimizer q h H(div, Ω) u 1 h α = q h λ h 1 u1 h, β = uh 1 + div q h u1 h, C F = 1 λh 1 λ 1 1 ( α + 1 ) β λ 1 λ1 λ1 X 2 + αx + β λ h 1, where X = λ 1 λ1

24 Algorithm min i λ i λ h 1 λ i C F ( u h 1 q h u h 1 Theorem: w u q + C F λ u + div q λ h 1 + C uh 1 + div q ) h F u1 h q H(div, Ω). Compute Rayleigh Ritz approximations λ h 1 and uh 1 Set λ = λ h 1 and u = u1 h Find approximate minimizer q h H(div, Ω) u 1 h α = q h λ h 1 u1 h, β = uh 1 + div q h u1 h, C F = 1 λh 1 λ 1 1 ( α + 1 ) β λ 1 λ1 λ1 λ1 X 2 + αx + β λ h 1, where X = λ 1 ) X2 2 λ 1, where X 2 = ( α 2 + 4(λ h1 β) α /2

25 Algorithm min i λ i λ h 1 λ i C F ( u h 1 q h u h 1 Theorem: w u q + C F λ u + div q λ h 1 + C uh 1 + div q ) h F u1 h q H(div, Ω). Compute Rayleigh Ritz approximations λ h 1 and uh 1 Set λ = λ h 1 and u = u1 h Find approximate minimizer q h H(div, Ω) u 1 h α = q h λ h 1 u1 h, β = uh 1 + div q h u1 h, C F = 1 λh 1 λ 1 1 ( α + 1 ) β λ 1 λ1 λ1 λ1 X 2 + αx + β λ h 1, where X = λ 1 ) X2 2 λ 1, where X 2 = ( α 2 + 4(λ h1 β) α /2 C F 1/X 2

26 Computing q h H(div, Ω) ( u q + C F λ u + div q ) 2 ) 2 ( u 1 h q + (λ h 1) 1/2 λ h 1u1 h + div q 1 + ϱ ϱ uh 1 q ϱ λ h λ h 1u1 h div q 2, ϱ > 1 Minimize over W h H(div, Ω): Find q h W h : (div q h, div ψ h ) + λh 1 ϱ (q h, ψ h ) = λh 1 ϱ ( uh 1, ψ h ) (λ h 1u h 1, div ψ h ) Solve by standard Raviart-Thomas finite elements. ψ h W h

27 Example 1 u = f in (, 2) (, 1) Γ N u = on Γ D n u = on Γ N Γ D f = 5π2 16 u u = sin πx 1 4 sin πx 2 2

28 Example 1 u = f in (, 2) (, 1) Γ N u = on Γ D n u = on Γ N f = 5π2 16 u u = sin πx 1 4 sin πx 2 2 C F = 4 5π. =.5694 C low C up F =.5693 F = Γ D Friedrichs constant Example 1.4 Upper bound p=1.2 Upper bound p=2 Exact value Lower bound Number of elements

29 Example 1 u = f in (, 2) (, 1) Γ N u = on Γ D n u = on Γ N Γ D Lower bound: reference solution Upper bound: error majorant 1 Error bounds Example Upper bound p=1 Upper bound p=2 Lower bound Number of elements

30 Example 2 u = f in (, 2) (, 1) u = on Γ D n u = on Γ N Γ D Γ N f = 5π2 16 sin πx 1 4 sin πx 2 2

31 Example 2 u = f in (, 2) (, 1) u = on Γ D n u = on Γ N Γ D Γ N f = 5π2 16 sin πx 1 4 sin πx 2 2 C F =? C low C up F =.775 F = Friedrichs constant Example 2.5 Upper bound p=1.25 Upper bound p=2 Lower bound Number of elements

32 Example 2 u = f in (, 2) (, 1) u = on Γ D n u = on Γ N Γ D Γ N Lower bound: reference solution Upper bound: error majorant 1 Error bounds Example 2 Upper bound p=1 Upper bound p=2 Lower bound Number of elements

33 Conclusions Practical method Guaranteed upper bound on Friedrichs constant Easy to generalize to similar inequalities Computationally demanding Exact representaion of the domain Ω curved elements Splines!

34 Thank you for your attention Tomáš Vejchodský Institute of Mathematics, Žitná 25, Praha 1 February 8, 212, SIGA 212, Prague