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.
Outline 1. Anisotropic Sobolev spaces 2. Boundary integral operators and equations 3. Boundary element methods 4. Preconditioning 5. FEM-BEM coupling 2
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
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
Outline 1. Anisotropic Sobolev spaces 2. Boundary integral operators and equations 3. Boundary element methods 4. Preconditioning 5. FEM-BEM coupling 5
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
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
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
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
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
Outline 1. Anisotropic Sobolev spaces 2. Boundary integral operators and equations 3. Boundary element methods 4. Preconditioning 5. FEM-BEM coupling 11
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
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, 1 1 4 (Σ) H 2, 1 4 (Σ). [γ 0 Ṽ w] = γ0 ext (Ṽ w) γint 0 (Ṽ w) = 0. 13
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
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
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
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
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
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
Outline 1. Anisotropic Sobolev spaces 2. Boundary integral operators and equations 3. Boundary element methods 4. Preconditioning 5. FEM-BEM coupling 20
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
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
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
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
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
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
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
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
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
Outline 1. Anisotropic Sobolev spaces 2. Boundary integral operators and equations 3. Boundary element methods 4. Preconditioning 5. FEM-BEM coupling 30
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
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 1 5 0 t 1 t 2 t 3 t 4 t 5 0 0.2 0.4 0.6 0.8 1 t 32
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. 0 2 2,249 0 1,001 1 1,002 1 1 4 1,311 0,778 2,808 2 1,279 2 2 8 0,658 0,996 4,905 4 1,422 4 3 16 0,324 1,021 7,548 8 1,486 8 4 32 0,16 1,017 11,14 16 1,541 14 5 64 0,079 1,01 16,724 31 1,563 13 6 128 0,04 1,006 13,47 41 1,59 13 7 256 0,02 1,003 22,053 50 1,615 12 8 512 0,01 1,001 32,043 59 1,636 12 9 1024 0,005 1,001 60,957 70 1,777 11 10 2048 0,002 1,000 88,488 82 1,762 11 11 4096 0,001 1,000 125,957 96 1,765 10 33
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. 15 10 x = 0 x = 1 wh(, t) 5 0 0 0.2 0.4 0.6 0.8 1 t 34
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. 0 2 1,886 1,001 2 1,001 2 1,002 2 1 3 1,637 3,972 3 2,553 3 1,16 3 2 5 1,272 12,225 5 4,055 4 1,166 4 3 7 0,914 34,212 7 3,611 6 1,156 6 4 9 0,615 92,081 9 3,164 8 1,149 8 5 11 0,401 118,586 11 2,945 10 1,224 10 6 13 0,267 338,26 13 2,803 12 1,21 12 7 20 0,166 621,773 20 3,524 18 1,197 13 8 31 0,101 1608,08 31 4,457 27 1,252 12 9 47 0,063 2344,9 47 5,779 32 1,574 11 10 74 0,039 6141,47 74 8,348 37 1,692 11 11 114 0,024 8409,92 114 10,95 42 1,561 10 12 177 0,015 23007,6 173 14,324 47 1,716 10 13 278 0,01 27528,3 200 21,094 53 1,677 10 35
Outline 1. Anisotropic Sobolev spaces 2. Boundary integral operators and equations 3. Boundary element methods 4. Preconditioning 5. FEM-BEM coupling 36
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
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
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
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
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
Sample triangulation 1D 1 0.8 0.6 T h E h Nodes I 0 Nodes I I t 0.4 0.2 0 0 0.2 0.4 0.6 0.8 1 x 42
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
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
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
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 0 8 4 0.031618 0 1 32 8 0.015779 1.003 2 128 16 0.004223 1.902 3 512 32 0.001119 1.915 4 2048 64 0.000290 1.951 5 8192 128 0.000074 1.974 46
Outlook Space-time error estimator for anisotropic BEM Ω R 3 Implementation... 47