Preconditioned space-time boundary element methods for the heat equation

Transcription

1 W I S S E N T E C H N I K L E I D E N S C H A F T Preconditioned space-time boundary element methods for the heat equation S. Dohr and O. Steinbach Institut für Numerische Mathematik Space-Time Methods for PDEs. RICAM.

2 Outline 1. Anisotropic Sobolev spaces 2. Boundary integral operators and equations 3. Boundary element methods 4. Preconditioning 5. FEM-BEM coupling 2

3 Model problem Dirichlet boundary value problem (Ω R n bounded Lipschitz-domain, Γ := Ω) α t u(x, t) x u(x, t) = 0 for (x, t) Q := Ω (0, T ), u(x, t) = g(x, t) for (x, t) Σ := Γ (0, T ), u(x, 0) = u 0 (x) for x Ω. Initial condition u 0 and boundary datum g given. Heat capacity α > 0. 3

4 Representation formula Representation formula for (x, t) Ω (0, T ) u(x, t) = Ω 1 α U (x y, t)u(y, 0)dy + 1 α T 0 Γ T n y U (x y, t s)u(y, s)ds yds. 0 Γ U (x y, t s) n y u(y, s)ds yds Fundamental solution ( ) n/2 ( ) α α x y 2 U exp, t > s, (x y, t s) := 4π(t s) 4(t s) 0, t s. 4

5 Outline 1. Anisotropic Sobolev spaces 2. Boundary integral operators and equations 3. Boundary element methods 4. Preconditioning 5. FEM-BEM coupling 5

6 Anisotropic Sobolev spaces For r, s 0 H r,s (R n R) := L 2 (R; H r (R n )) H s (R; L 2 (R n )) where u H s (R, L 2 (R n )) ( 1 + τ 2) s/2 û L 2 (R, L 2 (R n )). For Q = Ω (0, T ) H r,s (Q) := { u Q : u H r,s (R n R) }. 6

7 Anisotropic Sobolev spaces on Σ For r, s 0 H r,s (Σ) := L 2 (0, T ; H r (Γ)) H s (0, T ; L 2 (Γ)). Equivalent norm for 0 < r, s < 1 T u 2 H r,s (Σ) = u 2 L 2 (Σ) + Subspace and 0 Γ Γ u(x, t) u(y, t) 2 x y n 1+2r ds y ds x dt T T u(, t) u(, τ) 2 L + 2 (Γ) 0 0 t τ 1+2s dτdt. H r,s,0 (Σ) := L2 (0, T ; H r (Γ)) H s 0 (0, T ; L2 (Γ)) H r, s (Σ) := [ H r,s,0 (Σ) ]. 7

8 Dirichlet trace operator Theorem (Lions, Magenes 1972) Let r > 1 2 and s 0. Then there exists a linear and bounded operator γ0 int : H r,s (Q) H µ,ν (Σ) with γ int 0 u H µ,ν (Σ) c T u H r,s (Q) for all u H r,s (Q) where µ = r 1 2, ν = s s 2r and γ int 0 is an extension of γ int 0 u = u Σ for u C(Q). For r = 1 and s = 1 2 we have γint 0 : H 1, 1 2 (Q) H 1 2, 1 4 (Σ). 8

9 Neumann trace operator Theorem (Costabel 1990) The mapping γ int 1 : H 1, 1 2 (Q, α t x ) H 1 2, 1 4 (Σ) is linear and continuous. If u C 1 (Q) then γ 1 u = n u Σ. Subspace H 1, 1 2 (Q, α t x ) := { } u H 1, 1 2 (Q) : (α t x )u L 2 (Q). 9

10 An existence theorem Let u 0 L 2 (Ω), g H 1 2, 1 4 (Σ). The initial boundary value problem α t u x u = 0 in Q, u = g on Σ, u = u 0 on Ω {0} has a unique solution u H 1, 1 2 (Q, α t x ). Dirichlet trace γ int 0 u H 1 2, 1 4 (Σ) and Neumann trace γ int 1 u H 1 2, 1 4 (Σ) well defined. 10

11 Outline 1. Anisotropic Sobolev spaces 2. Boundary integral operators and equations 3. Boundary element methods 4. Preconditioning 5. FEM-BEM coupling 11

12 Initial potential Let u 0 L 2 (Ω). The function ( M 0 u 0 )(x, t) := U (x y, t)u 0 (y)dy is a solution of the homogeneous heat equation for (x, t) Q. Ω The operator M 0 : L 2 (Ω) H 1, 1 2 (Q, α t x ) is linear and bounded. boundedness of M 0 := γ0 int M 0 : L 2 (Ω) H 1 2, 1 4 (Σ) M 1 := γ1 int M 0 : L 2 (Ω) H 1 2, 1 4 (Σ). 12

13 Single layer potential Single layer potential with density w (Ṽ w)(x, t) := 1 t U (x y, t s)w(y, s)ds y ds α 0 Γ is a solution of the homogeneous heat equation for (x, t) Q. The operator Ṽ : H 1 2, 1 4 (Σ) H 1, 1 2 (Q, α t x ) is linear and bounded. boundedness of the single layer boundary integral operator Jump relation V := γ0 int Ṽ : H 1 2, (Σ) H 2, 1 4 (Σ). [γ 0 Ṽ w] = γ0 ext (Ṽ w) γint 0 (Ṽ w) = 0. 13

14 Adjoint double layer potential Adjoint double layer potential with density w and (x, t) Σ (K w)(x, t) := 1 α t 0 Γ n x U (x y, t s)w(y, s)ds y ds. The operator K : H 1 2, 1 4 (Σ) H 1 2, 1 4 (Σ) is linear and bounded and γ int 1 (Ṽ w) = 1 2 w + K w. Jump relation [γ 1 Ṽ w] = γ1 ext (Ṽ w) γint 1 (Ṽ w) = w. 14

15 Double layer potential Double layer potential with density v (Wv)(x, t) := 1 α t 0 Γ n y U (x y, t s)v(y, s)ds y ds is a solution of the homogeneous heat equation for (x, t) Q The operator W : H 1 2, 1 4 (Σ) H 1, 1 2 (Q, α t x ) is linear and bounded. boundedness of γ int 0 W : H 1 2, 1 4 (Σ) H 1 2, 1 4 (Σ). 15

16 Double layer boundary integral operator with density v and (x, t) Σ (Kv)(x, t) := 1 α t 0 Γ n y U (x y, t s)v(y, s)ds y ds. The operator K : H 1 2, 1 4 (Σ) H 1 2, 1 4 (Σ) is linear and bounded and Jump relation γ int 0 (Wv) = 1 2 v + Kv. [γ 0 Wv] = γ ext 0 (Wv) γ int 0 (Wv) = v. 16

17 Hypersingular boundary integral operator Hypersingular boundary integral operator with density v and (x, t) Σ (Dv)(x, t) := γ int 1 (Wv)(x, t). The operator D : H 1 2, 1 4 (Σ) H 1 2, 1 4 (Σ) is linear and bounded. Jump relation [γ 1 Wv] = γ ext 1 (Wv) γ int 1 (Wv) = 0. 17

18 Boundary integral equations Representation formula for ( x, t) Q u( x, t) = (Ṽ γint 1 u)( x, t) (W γ int 0 u)( x, t) + ( M 0 u 0 )( x, t). Apply Dirichlet- and Neumann trace operators ( γ int 0 u ) ( 1 γ1 intu = 2 I K V ) ( γ int 1 D 2 I + K 0 u ) γ1 int }{{} u =: C Calderón-operator: C = C 2. ( M0 u + 0 M 1 u 0 ). 18

19 Theorem (Costabel 1990) There exists a constant c 1 > 0, such that ( ( ) ( ψ V K ψ ( ), ϕ) K c D ϕ) 1 ψ 2 + H 1 2, 4 1 (Σ) ϕ 2 H 1 2, 4 1 (Σ) for all (ψ, ϕ) H 1 2, 1 4 (Σ) H 1 2, 1 4 (Σ). Ellipticity of V and D, i.e. and Vw, w c V 1 w 2 H 1 2, 1 4 (Σ) Dv, v c D 1 v 2 H 1 2, 1 4 (Σ) for all w H 1 2, 1 4 (Σ) for all v H 1 2, 1 4 (Σ). 19

20 Outline 1. Anisotropic Sobolev spaces 2. Boundary integral operators and equations 3. Boundary element methods 4. Preconditioning 5. FEM-BEM coupling 20

21 Boundary element methods First boundary integral equation for (x, t) Σ γ int 0 u(x, t) = (V γ int 1 u)(x, t)+ 1 2 γint 0 u(x, t) (K γ int 0 u)(x, t)+(m 0 u 0 )(x, t). Direct approach: Find γ1 intu H 1 2, 1 4 (Σ), such that ( ) 1 V γ1 int u = 2 I + K g M 0 u 0 on Σ. Variational formulation is to find γ1 intu H 1 2, 1 4 (Σ), such that ( ) 1 V γ1 int u, τ Σ = 2 I + K g, τ Σ M 0 u 0, τ Σ for all τ H 1 2, 1 4 (Σ). Uniquely solvable due to ellipticity and boundedness of V. 21

22 Indirect approach: u( x, t) := (Ṽ w)( x, t) + ( M 0 u 0 )( x, t) for ( x, t) Q. Apply Dirichlet trace operator g(x, t) = (Vw)(x, t) + (M 0 u 0 )(x, t) for (x, t) Σ. Variational formulation is to find w H 1 2, 1 4 (Σ), such that Vw, τ Σ = g M 0 u 0, τ Σ for all τ H 1 2, 1 4 (Σ). Uniquely solvable due to ellipticity and boundedness of V. 22

23 Triangulation of Σ for n = 1, 2 Boundary Γ piecewise smooth with Γ = J j=1 Γ j. Σ = J j=1 Σ j with Σ j := Γ j (0, T ). Σ N = N l=1 σ l admissible decomposition of Σ. For each σ l exists j: σ l Σ j. σ l = χ l (σ) with reference element σ R n. Ω (0, T ) Assumptions: No curved elements. Boundary elements are shape regular. 23

24 Trial spaces Space of piecewise constant basis functions S 0 h (Σ) := span { ϕ 0 l } N l=1 with ϕ 0 l (x, t) := { 1 for (x, t) σ l, 0 else. Approximate w := γ int 1 u H 1 2, 1 4 (Σ) by w h (x, t) := N w l ϕ 0 l (x, t) S0 h (Σ). l=1 24

25 Galerkin-Bubnov variational formulation: Find w h Sh 0 (Σ), such that ( ) 1 Vw h, τ h Σ = 2 I + K g, τ h Σ M 0 u 0, τ h Σ for all τ h Sh 0 (Σ). Equivalent to with and for l, k = 1,..., N. V h w = f V h [l, k] = V ϕ 0 k, ϕ0 l Σ ( ) 1 f [l] = 2 I + K g, ϕ 0 l Σ M 0 u 0, ϕ 0 l Σ 25

26 Approximation properties For u H r,s (Σ) with r, s [0, 1] and for σ, µ [ 1, 0) u Q h u L2 (Σ) c (hr + h s t ) u H r,s (Σ) u Q h u H σ,µ (Σ) c ( h σ + h µ t ) (h r + h s t ) u H r,s (Σ). For n = 1: Terms with h vanish. For n = 2: h t denotes the size of an element in temporal direction, h the size in spatial direction. 26

27 Space of piecewise smooth functions For Σ j = Γ j (0, T ) and r, s 0 H r,s (Σ j ) := { v = ṽ Σj : ṽ H r,s (Σ) }. Space of piecewise smooth functions on Σ H r,s pw (Σ) := { v L 2 (Σ) : v Σj H r,s (Σ j ) for j = 1,..., J } with norm v H r,s pw (Σ) := J v Σj 2 j=1 H r,s (Σ j ) 1/2. 27

28 For r, s < 0 H r,s (Σ j ) := [ H r, s (Σ j ) ] and with norm H r,s pw (Σ) := w H r,s pw (Σ) := J H r,s (Σ j ) j=1 J Hr,s w Σj. (Σ j ) j=1 For w H r,s pw (Σ) with r, s < 0 we have w H r,s (Σ) w H r,s pw (Σ). 28

29 Error estimates Quasi-optimality of the solution w h S 0 h (Σ) w w h H 1 2, 1 4 (Σ) c J inf j=1 τ j h S0 h (Σ j ) w Σj τ j h H 1 2, 1 4 (Σ j ) For w Hpw r,s (Σ) with r, s [0, 1] ( ) w w h c H h 1/2 + h 1/4 1 2, 1 4 (Σ) t (h r + ht s ) w H r,s pw (Σ).. For n = 1 w w h L 2 (Σ) chs t w H s pw (Σ). 29

30 Outline 1. Anisotropic Sobolev spaces 2. Boundary integral operators and equations 3. Boundary element methods 4. Preconditioning 5. FEM-BEM coupling 30

31 Preconditioning V and D elliptic inf-sup condition. Find subspaces X h = span {ϕ k } N k=1 H 1 2, 1 4 (Σ) and Y h = span {ψ l } M l=1 H 1 2, 1 4 (Σ), such that (τ h, v h ) sup c1 M τ h 0 v h Y h v h 1 H, H 1 1 2, 1 4 (Σ) 2 4 (Σ) for all τ h X h and dimx h = dimy h. ( κ M 1 h ) D hm T h V h c where M h [l, k] = (ϕ k, ψ l ) for l, k = 1,..., N. 31

32 Different approaches for n = 1: Sh 0 (I) for V and D. Sh 1 (I) for V and D. Use dual mesh: Sh 1(I) for D and S0 h (Ĩ) corresponding to dual mesh for V (figure: Sample dual mesh for n = 1). 1 ϕ 0 1 ϕ 0 2 ϕ 0 3 ϕ 0 4 ϕ 0 5 ϕ 1 1 ϕ 1 2 ϕ 1 3 ϕ 1 4 ϕ t 1 t 2 t 3 t 4 t t 32

33 Numerical examples Uniform refinement. Ω = (0, 1), T = 1. Initial condition u 0 (x) = sin (2πx). Boundary condition g = 0. Sh 0 (I) for the discretization of V and D. L N w w h L2 (Σ) eoc κ(v h ) It. κ(c 1 V V h) It , , , ,311 0,778 2, , ,658 0,996 4, , ,324 1,021 7, , ,16 1,017 11, , ,079 1,01 16, , ,04 1,006 13, , ,02 1,003 22, , ,01 1,001 32, , ,005 1,001 60, , ,002 1,000 88, , ,001 1, , ,

34 Adaptive refinement. Ω = (0, 1), T = 1. Initial condition u 0 (x) = 5 exp ( 10t) sin (πx). Boundary condition g = 0. Sh 0 (I) for the discretization of V and D x = 0 x = 1 wh(, t) t 34

35 C V = diagv h C V = M h D 1 h M h L N w w h L2 (Σ) κ(v h ) It. κ(c 1 V V h) It. κ(c 1 V V h) It ,886 1, , , ,637 3, , , ,272 12, , , ,914 34, , , ,615 92, , , , , , , , , , , , , , , , , , , , ,9 47 5, , , , , , , , , , , , , , , , , ,

36 Outline 1. Anisotropic Sobolev spaces 2. Boundary integral operators and equations 3. Boundary element methods 4. Preconditioning 5. FEM-BEM coupling 36

37 FEM-BEM coupling Transmission problem α t u i (x, t) div x [A(x, t) x u i (x, t)] = f (x, t) for (x, t) Ω (0, T ), α t u e (x, t) u e (x, t) = 0 for (x, t) Ω ext (0, T ), u i (x, 0) = u 0 (x) for x Ω, u e (x, 0) = 0 where f L 2 (0, T ; H 1 (Ω)), u 0 H 1 0 (Ω). Transmission conditions for (x, t) Σ for x Ω ext u i (x, t) = u e (x, t), n x [A(x, t) x u i (x, t)] = n x u e (x, t) =: w e (x, t). Radiation condition for x. 37

38 Representation formula for ( x, t) Ω ext (0, T ) u e ( x, t) = 1 α + 1 α T 0 Γ T 0 Γ n y u e (y, s)u ( x y, t s)ds y ds u e (y, s) n y U ( x y, t s)ds y ds. Apply Dirichlet-trace operator γ ext 0 u e = V γ ext 1 u e + ( ) 1 2 I + K γ0 ext u e on Σ. 38

39 Variational formulation Consider decomposition u i (x, t) = u i (x, t) + u 0 (x, t) for (x, t) Q where u 0 L 2 (0, T ; H 1 0 (Ω)) H1 (0, T ; H 1 (Ω)) is an extension of u 0. Find u i L 2 (0, T ; H 1 (Ω)) H 1 (0, T ; H 1 (Ω)) with u i (x, 0) = 0 for x Ω, such that for all v L 2 (0, T ; H 1 (Ω)). Bilinear form a(u, v) := α Q a(u i, v) w i, v Σ = f, v Q a(u 0, v) t u(x, t)v(x, t)dxdt+ [A(x, t) x u(x, t)] x v(x, t)dxdt. Q 39

40 Variational formulation BIE: Find w e X, such that ( ) 1 Vw e, τ Σ + 2 I K u e, τ Σ = 0 for all τ X. Transmission conditions u i Σ = u i Σ = u e Σ and w i = w e Find u i L 2 (0, T ; H 1 (Ω)) H 1 (0, T ; H 1 (Ω)) with u i (x, 0) = 0 for x Ω and w e X, such that a(u i, v) w e, v Σ = f, v Q a(u 0, v), ( ) 1 Vw e, τ Σ + 2 I K u i, τ Σ = 0 for all v L 2 (0, T ; H 1 (Ω)) and τ X. 40

41 Triangulation Admissible triangulation T h = {T k } N Q k=1 of Q. M Q... Number of nodes I 0... Index set of nodes not belonging to Ω {0} M 0 := I 0 I I... Index set of nodes not belonging to Σ (Ω {0}) M I := I I Node sorting: I I {1,..., M I }, I 0 {1,..., M 0 }. Boundary elements E h = {σ k } N Σ k=1 given by E h := { σ Σ : T T h : σ = T Σ }. 41

42 Sample triangulation 1D T h E h Nodes I 0 Nodes I I t x 42

43 Trial spaces Sh 0(Σ) = span { ϕ 0 k functions. Sh 1(Q) = span { ϕ 1 i basis functions. } NΣ k=1 } MQ i=1 space of piecewise constant basis space of piecewise linear and continuous Sh,0 1 (Q) functions in S1 h (Q) vanishing on Ω {0}, i.e. Sh,0 1 (Q) = span { } ϕ 1 M0 i i=1. Approximate w e and u i by N Σ w e,h = w k ϕ 0 k S0 h (Σ), u M 0 i,h = u j ϕ 1 j k=1 j=1 S 1 h,0 (Q). 43

44 Galerkin variational formulation Let u 0,h be the interpolation of u 0 in S 1 h (Q). Find u i,h Sh,0 1 (Q) and w e,h Sh 0 (Σ), such that a(u i,h, v h ) w e,h, v h Σ = f, v h Q a(u 0,h, v h ), ( ) 1 Vw e,h, τ h Σ + 2 I K u i,h, τ h Σ = 0 for all v h S 1 h,0 (Q) and τ h S 0 h (Σ). 44

45 Equivalent system of linear equations with A QQ A QΣ A ΣQ A ΣΣ M T h 1 2 M h K h V h uq u Σ w = f Q f Σ 0 A[j, i] = a(ϕ 1 i, ϕ1 j ), M Q f [j] = f, ϕ1 j Q u0 r a(ϕ 1 r, ϕ 1 j ) r=1 for i, j = 1,..., M 0 and M h [l, i] = ϕ 1 M I +i, ϕ0 l Σ, K h [l, i] = K ϕ 1 M I +i, ϕ0 l Σ, V h [l, k] = V ϕ 0 k, ϕ0 l Σ for i = 1,..., M 0 M I and k, l = 1,..., N Σ. 45

46 Numerical example Ω = (0, 1), T = 1, A 1, f 0, initial condition ( ) 1 exp u 0 (x) = (2x 1) 2 sin (πx) for x (0, 1), 1 0 else. L N Q N Σ u i u L i,h 2 (Q) eoc

47 Outlook Space-time error estimator for anisotropic BEM Ω R 3 Implementation... 47