Space time finite and boundary element methods

Transcription

1 Space time finite and boundary element methods Olaf Steinbach Institut für Numerische Mathematik, TU Graz based on joint work with M. Neumüller, H. Yang, M. Fleischhacker, S. Bonkhoff, S. Dohr, K. Niino, M. Zank 1 / 29

2 Heat equation for (x,t) Q := (,1) (,T) αu t (x,t) u xx (x,t) =, u(,t) = u(1,t) =, u(x,) = u (x) Solution u(x,t) = u k e (kπ)2t/α sinkπx, u k = 2 1 k=1 u (x)sinkπx dx Uniform/adaptive piecewise linear finite element interpolation in Q level nodes u I h u L 2 (Q) eoc / 29

3 Solution of the heat equation and normal derivatives at boundary nodes Solution of the wave equation and normal derivatives at boundary nodes. 3 / 29

4 Dirichlet boundary value problem α t u(x,t) x u(x,t) = f(x,t) for (x,t) Q := (,T), u(x,t) = g(x,t) for (x,t) Σ := Γ (,T), u(x,) = u (x) for x. Compatibility condition: g(x,) = u (x) for x Γ. Variational formulation Find u L 2 (,T;H 1 ()) H 1 (,T,H 1 ()), u(x,t) = g(x,t) for x Γ, t (,T), u(x,) = u (x) for x, such that T [ ] α t u(x,t)v(x,t)+ x u(x,t) x v(x,t) dx dt is satisfied for all v L 2 (,T;H 1 ()). = T f(x,t)v(x,t)dx dt 4 / 29

5 Bilinear form a (u,v) := T [ ] α t u(x,t)v(x,t)+ x u(x,t) x v(x,t) dx dt Consider u p L 2 (,T;H 1 ()) H 1 (,T;H 1 ()) satisfying u p (x,t) = g(x,t) for (x,t) Σ, u p (x,) = u (x) for x Primal Formulation: Find u L 2 (,T;H 1 ()) H1, (,T;H 1 ()) such that is satisfied for all v L 2 (,T;H 1 ()). a (u +u p,v) = f,v Q Galerkin Petrov variational formulation; unique solvability based on inf sup stability condition [Schwab, Stevenson 29; Urban, Patera 214; Andreev 213; Mollet 214;...] 5 / 29

6 Find u L 2 (,T;H 1()) H1, (,T;H 1 ()) such that a (u,v) = f,v Q a (u p,v) is satisfied for all v L 2 (,T;H 1()). Operator equation to find u L 2 (,T;H 1()) H1, (,T;H 1 ()) such that Au = f in L 2 (,T;H 1 ()) i.e. u L 2 (,T;H()) H 1,(,T;H 1 1 ()) : Au f L 2 (,T;H 1 ()) = 6 / 29

7 For u L 2 (,T;H 1 ()) H 1 (,T;H 1 ()) find w L 2 (,T;H 1 ()) such that x w(x,t) = α t u(x,t) x u(x,t) in Q = (,T), w(x,t) = Find w L 2 (,T;H 1 ()) such that T is satisfied for all v L 2 (,T;H 1 ()) Lemma on Σ = Γ (,T). x w(x,t) x v(x,t)dx dt = a (u,v) α t u x u 2 L 2 (,T;H 1 ()) = w 2 L 2 (,T;H 1 ()) = a (u,w) 7 / 29

8 Theorem: For u L 2 (,T;H 1 ()) H1, (,T;H 1 ()) we have α t u x u L 2 (,T;H 1 ()) sup v L 2 (,T;H 1 ()) a (u,v) v L 2 (,T;H 1 ()) Lemma: Norm equivalence for u X = L 2 (,T;H 1 ()) H1, (,T;H 1 ()) u 2 X = α t u x u 2 L 2 (,T;H 1 ()) α tu 2 L 2 (,T;H 1 ()) + u 2 L 2 (,T;H 1 ()) Finite element discretization schemes space time tensor product wavelets [Schwab, Stevenson 29; Urban, Patera 214; Andreev 213; Mollet 214] Space time DG FEM [Neumüller 213] Space time FEM [OS 215] 8 / 29

9 Finite element spaces X h X = L 2 (,T;H()) H 1,(,T;H 1 1 ()), Y h Y = L 2 (,T;H()) 1 Assumption X h Y h (X h = Y h ) Galerkin Petrov variational formulation Norms Find u h X h : a (u h,v h ) = f,v h Q a (u p,v h ) for all v h Y h u 2 X = α t u x u 2 L 2 (,T;H 1 ()) α t u 2 L 2 (,T;H 1 ()) + u 2 L 2 (,T;H 1()) = w 2 L 2 (,T;H 1()) + u 2 L 2 (,T;H 1()) where w Y solves T x w(x,t) x v(x,t)dx dt = T for all v Y. α t u(x,t)v(x,t)dx dt 9 / 29

10 Define w h Y h as unique solution of T x w h (x,t) x v h (x,t)dx dt = T for all v h Y h. Discrete norm α t u(x,t)v h (x,t)dx dt u 2 X h := w 2 L 2 (,T;H 1 ()) + u 2 L 2 (,T;H 1 ()) u 2 X Theorem Theorem (Cea s Lemma) u a(u h,v h ) h Xh sup v h Y h v h Y for all u h X h Theorem (Approximation property) u u h Xh 5 inf z h X h u z h X u u h Xh ch s 1 u Hs (Q), s p +1, u H s (Q) 1 / 29

11 Numerical example u(x,t) = cosπtsinπx for (x,t) Q = (,1) 2 p = 1 p = 2 L N dof u u h L 2 (,T;H 1()) eoc dof u u h L 2 (,T;H 1()) eoc / 29

12 Numerical example u(x,t) = (1 t) α sinπx for (x,t) Q = (,1) 2, α =.75, p = 1 p = 2 L N dof u u h L 2 (,T;H 1()) eoc dof u u h L 2 (,T;H 1()) eoc reduced regularity: u H 5/4 ε (Q),ε > eoc:.25? but optimal convergence for pw linear finite elements, eoc 1. eoc for pw quadratic finite elements: / 29

13 Lemma Let Q h R 2 be a space time finite element mesh of right angled isosceles triangles. Let u be a given function such that x u H 1 (Q) is satisfied, and let I h u be the continuous piecewise linear interpolation. Then there holds the error estimate Reduced regularity u I h u L2 (,T;H 1 ()) ch x u H1 (Q). x u H 1 (Q) instead of u H 2 (Q) Conjecture: The result remains valid for arbitrary shape regular finite elements spatial dimensions n = 2,3 Question: Interpolation error estimates for second order elements? 13 / 29

14 Institut fu r Numerische Mathematik A posteriori error indicators η2 = N P ℓ=1 kwℓ k2h 1 (qℓ ), [H. Yang; M. Fleischhacker; R. Verfu hrt] wℓ + wℓ = nx wℓ wℓ = = f α t uh + uh 1 2 [nx x uh ] in qℓ on qℓ \Σ on qℓ Σ t Example x Solution O. Steinbach Interpolation FEM RICAM 216, Linz, / 29

15 Primal Formulation Find u L 2 (,T;H 1 ()) H 1, (,T;H 1 ()), u(x,t) = g(x,t) for x Γ, t (,T), such that T [ ] α t u(x,t)v(x,t)+ x u(x,t) x v(x,t) dx dt = is satisfied for all v L 2 (,T;H 1 ()). Dual variational formulation T f(x,t)v(x,t)dx dt Find u L 2 (,T;H 1 ()), u(x,t) = g(x,t) for x Γ, t (,T), such that T [ ] u(x,t)α t v(x,t)+ x u(x,t) x v(x,t) dx dt = T is satisfied for all v L 2 (,T;H 1 ()) H1, (,T;H 1 ()). f(x,t)v(x,t)dx dt + αu (x)v(x,)dx 15 / 29

16 Dirichlet boundary value problem α t u(x,t) x u(x,t) = f(x,t) for (x,t) Q := (,T), u(x,t) = for (x,t) Σ := Γ (,T), u(x,) = for x. Primal variational formulation L : L 2 (,T;H()) H 1,(,T;H 1 1 ()) [L 2 (,T;H())] 1 Dual variational formulation L : L 2 (,T;H()) 1 [L 2 (,T;H()) H 1,(,T;H 1 1 ())] Interpolation, θ [, 1] [Lions, Magenes 1972] L : [L 2 (,T;H()) H 1,(,T;H 1 1 ()),L 2 (,T;H())] 1 θ [[L 2 (,T;H())] 1,[L 2 (,T;H()) H 1,(,T;H 1 1 ())] ] θ = [L 2 (,T;H()) H 1,(,T;H 1 1 ()),L 2 (,T;H())] 1 1 θ θ = 1 2 L : H 1,1 2 (Q) [H 1,1 2 (Q)] 16 / 29

17 Dirichlet trace operator [Lions, Magenes 1972] γ : H 1,1 2 (Q) H 1 2,1 4 (Σ) Analysis of boundary integral operators D. N. Arnold, P. J. Noon 1987, 1989; P. J. Noon 1988 G. C. Hsiao, J. Saranen 1989, 1993 M. Costabel ellipticity of single layer boundary integral operator V : H 1 2, 1 4 (Σ) H 1 2,1 4 (Σ) + standard analysis of discretization schemes? results are based on interpolation between primal and dual formulation? link to Green s formula not obvious? coupling of finite and boundary element methods 17 / 29

18 Dirichlet boundary value problem α t u(x,t) x u(x,t) = for (x,t) Q := (,T), u(x,t) = g(x,t) for (x,t) Σ := Γ (,T), u(x,) = u (x) for x. Representation formula for x and t (,T) u(x,t) = 1 T U (x y,t s) ny u(y,s)ds y ds α Γ 1 T ny U (x y,t s)g(y,s)ds y ds + U (x y,t)u (y)dy α Γ Fundamental solution ( ) n/2 ) α α x y 2 exp( for s [,t), U (x y,t s) = 4π(t s) 4(t s) for s t. 18 / 29

19 Single layer potential Duality Adjoint problem (Ṽw)(x,t) = 1 α T Γ U (x y,t s) ny u(y,s)ds y ds Ṽw,ψ Q = w,ϕ Σ α s ϕ(y,s) y ϕ(y,s) = ψ(y,s) ϕ(y,t) = for y,s (,T), for y Primal formulation: ϕ L 2 (,T;H 1 ()) H, 1 (,T;H 1 ()) T [ ] α s ϕ(y,s)φ(y,s)+ y ϕ(y,s) y φ(y,s) dy ds = T Single layer potential ψ(y,s)φ(y,s)dy ds for all φ L 2 (,T;H 1 ()) Ṽ : [γ L 2 (,T;H 1 ()) H 1,(,T;H 1 ())] L 2 (,T;H 1 ()) 19 / 29

20 Dual formulation: ϕ L 2 (,T;H 1 ()) T = [ ] ϕα s φ(y,s)+ y ϕ(y,s) y φ(y,s) dy ds T Single layer potential Corollary ψ(y,s)φ(y,s)dy ds φ L 2 (,T;H 1 ()) H 1,(,T;H 1 ()) Ṽ : L 2 (,T;H 1/2 (Γ)) L 2 (,T;H 1 ()) H 1,(,T; H 1 ()) V : L 2 (,T;H 1/2 (Γ)) V Σ := γ [L 2 (,T;H 1 ()) H 1,(,T; H 1 ())] 2 / 29

21 Single layer potential Ṽ : L 2 (,T;H 1/2 (Γ)) L 2 (,T;H 1 ()) H 1,(,T; H 1 ()) Green s formula for u = Ṽw: T n u(x,t)v(x,t)ds x dt = a (u,v) Lemma Γ n u L2 (,T;H 1/2 (Γ)) c a 2 u L2 (,T;H 1 ()) H 1 (,T; H 1 ()) 21 / 29

22 Proof of stability: [Nedelec, Planchard 1973; Hsiao, Wendland 1977; Costabel 199] T T x u x v dx dt +α t uv dx dt = T n uv ds x dt + T Γ For u = Ṽw, u Γ = Vw, n u = ( 1 2 I +K )w we obtain [α t u x u]v dx dt a (u,v) = ( 1 2 I +K )w,v Σ Correspondingly, a c(u,v) = ( 1 2 I +K )w,v Σ Hence, a (u,v)+a c(u,v) = w,v Σ In particular for v = u this finally gives T w,vw Σ = x u 2 dxdt + α [u(t)] 2 dx R 2 n R n 22 / 29

23 Alternatively, for u = Ṽw, u Γ = Vw, n u = ( 1 2 I +K )w we may define x z = c H t u in R n \Γ, z Γ = and consider Hence, v = u +z = u +N t u, v Γ = u Γ = Vw w,vw Σ = a (u,u +N t u)+a c(u,u +N t u) 1 ( u 2 L 2 2(,T;H 1 ()) + tu 2 L 2(,T; H 1 ()) c V 1 w 2 L 2 (,T;H 1/2 (Γ)) Lemma: For w L 2 (,T;H 1/2 (Γ)) + u 2 L 2(,T;H 1 ( c )) + tu 2 L 2(,T; H 1 ( c )) Vw,φ Σ c S w L2 (,T;H 1/2 (Γ)) sup φ V Σ φ V Σ ) 23 / 29

24 Boundary element mesh { N 1 for (x,t) σl, Σ = σ l, ψ l (x,t) = else Example l=1 α = 1, u (x) = sinπx for x (,1), g(,t) = g(1,t) = L N BEM Iter w w h L 2 (,T) eoc / 29

25 Representation formula (g ) u(x,t) = 1 α T Approximate representation formula ũ(x,t) = 1 α T Γ Γ U (x y,t s)w(y,s)ds y ds + U (x y,t)u (y)dy U (x y,t s)w h (y,s)ds y ds + U (x y,t)u (y)dy Define piecewise linear finite element interpolation Local error indicators ũ h = I h ũ η k = ũ I h ũ L2 (q k ) Adaptive finite element boundary element solution [OS 1999] 25 / 29

26 Representation formula (g ) u(x,t) = 1 α T Normal derivative on Σ w(x,t) = 1 2 w h(x,t)+ 1 α Γ U (x y,t s)w(y,s)ds y ds + U (x y,t)u (y)dy T Error equation Laplace: [H. Schulz, OS 2] Γ w h (y,s) nx U (x y,t s)ds y ds + nx U (x y,t)u (y)dy ( 1 2 I K )(w w h ) = w w h on Σ Error indicator ẽ h (x,t) = w(x,t) w h (x,t) 26 / 29

27 L N BEM w w h L 2 (,T) eoc N FEM M FEM ũ ũ h L 2 (Q) eoc uniform refinement adaptive refinement / 29

28 Conclusions space time finite and boundary element methods arbitrary space time meshes, no time stepping schemes adapativity in space and time simultaneously a posteriori error estimators parallel iterative solvers and preconditioners Fast BEM / 29

29 3. Chemnitz FEM Symposium Strobl, September 25 27, Söllerhaus Workshop on Fast Boundary Element Methods in Industrial Applications October 12 15, / 29