FEM for Stokes Equations

Transcription

1 FEM for Stokes Equations Kanglin Chen 15. Dezember 2009

2 Outline 1 Saddle point problem 2 Mixed FEM for Stokes equations 3 Numerical Results 2 / 21

3 Stokes equations Given (f, g). Find (u, p) s.t. u + p = f in Ω, divu = g in Ω, u = 0 on Ω. 3 / 21

4 Stokes equations Given (f, g). Find (u, p) s.t. u + p = f in Ω, divu = g in Ω, u = 0 on Ω. The Stokes equations describe two- or three-dimensional viscous ow. If g = 0, the ow is incompressible. 3 / 21

5 Variational formulation Find (u, p) X Q s.t. { a(u, v) + b(v, p) = (f, v) v X, b(u, µ) = (g, µ) µ Q, (1) where a(u, v) := Ω (f, v) := u : vdxdy, b(v, p) := f vdxdy, (g, µ) := Ω divvpdxdy, gµdxdy. Ω Ω 4 / 21

6 Variational formulation Find (u, p) X Q s.t. { a(u, v) + b(v, p) = (f, v) v X, b(u, µ) = (g, µ) µ Q, (1) where a(u, v) := Ω (f, v) := u : vdxdy, b(v, p) := f vdxdy, (g, µ) := Ω divvpdxdy, gµdxdy. Ω Ω We call this the saddle point problem. Why? 4 / 21

7 Operator equations Dene a linear mapping L : X Q X Q (u, p) (f, g) where X := H 1 0 (Ω), Q := L2 0 (Ω) = { p L 2 (Ω) : pdxdy = 0 Ω }. 5 / 21

8 Operator equations Dene a linear mapping L : X Q X Q (u, p) (f, g) where X := H 1 0 (Ω), Q := L2 0 (Ω) = { For the bilinear form a we can nd operator A A : X X, (Au, v) = a(u, v) v X. Fachbereich 03 p L 2 (Ω) : pdxdy = 0 Ω }. 5 / 21

9 Operator equations Dene a linear mapping L : X Q X Q (u, p) (f, g) where X := H 1 0 (Ω), Q := L2 0 (Ω) = { For the bilinear form a we can nd operator A A : X X, (Au, v) = a(u, v) v X. For the bilinear form b we can nd operator B Fachbereich 03 p L 2 (Ω) : pdxdy = 0 Ω B : X Q B T : Q X, (Bu, µ) = b(u, µ) µ Q (B T p, v) = b(v, p) v X. }. 5 / 21

10 Operator equations The variational problem (1) is equivalent to { Au + B T p = f, Bu = g. 6 / 21

11 Operator equations The variational problem (1) is equivalent to { Au + B T p = f, Bu = g. Lemma The following statements are equivalent 1 There exists some β > 0 with inf sup b(v, µ) µ Q v X v µ β. 2 The operator B : V Q is an isomorphism. V := {v X : b(v, µ) = 0 µ Q}. 3 The operator { B T : Q V 0 is an} isomorphism. V 0 := l X : (l, v) = 0, v V. 6 / 21

12 Brezzis Splitting Theorem Brezzis Splitting Theorem The solution operator L : X Q X Q (u, p) (f, g) is an isomorphism i the following conditions are fullled 1 the bilinear form a is coercive, i.e. α > 0 s.t. a(v, v) α v 2, v V. 2 the bilinear form b fullls the inf-sup condition. 7 / 21

13 Discrete variational problem Find (u h, p h ) V h Q h such that { a(uh, v) + b(v, p h ) = (f, v) v V h X, b(u h, q) = 0, q Q h Q, 8 / 21

14 Discrete variational problem Find (u h, p h ) V h Q h such that { a(uh, v) + b(v, p h ) = (f, v) v V h X, b(u h, q) = 0, q Q h Q, Regarding existence and uniqueness of solutions, we have to specify adequate nite element subspaces V h and Q h such that the discrete inf-sup condition is fullled. β > 0 : sup v V h b(v, p h ) v p h β, p h Q h 8 / 21

15 P2-P1 Triangulation Not every nite element space satises the discrete inf-sup condition. The P2-P1 approximation, so-called Taylor and Hood elements in [1], works. 9 / 21

16 P2-P1 Triangulation Not every nite element space satises the discrete inf-sup condition. The P2-P1 approximation, so-called Taylor and Hood elements in [1], works. V h = Q h = {v C 0 ( Ω) 2 v k P 2 (Ω) 2, k T h } { } q C 0 ( Ω) q k P 1 (Ω), k T h. 9 / 21

17 Linear system Let (Φ i ), (ψ i ) be the basic functions of V h and Q h, then we have u h = a(u h, Φ j ) = b(φ i, p h ) = 2N M α i Φ i, p h = γ i ψ i i=1 i=1 2N α i a(φ i, Φ j ), A ij = a(φ i, Φ j ) i=1 M γ i b(φ i, ψ j ), Bij T = b(φ i, ψ j ) i=1 f 2N = (f i ), f i = (f, Φ i ) u 2N = {α i }, p M = {γ i }. Then we derive ( ) ( ) ( ) A B T u2n f2n =. B 0 p M 0 Fachbereich / 21

18 Linear system The stiness matrix A has the block form ( ) A1 0 A =, 0 A 1 where A 1 = ( Ω ϕ i ϕ j dxdy ) ij, i, j = 1,, N. The matrix BT has also the block form B T 1 = B T 2 = Ω Ω ϕ i x ψ j dxdy, ϕ i y ψ j dxdy, ( B B T T = 1 B T 2 ), i = 1,, N; j = 1,, M i = N + 1,, 2N; j = 1,, M. 11 / 21

19 Basic functions In element E the basic functions of V h are given by ϕ 1 (x, y) = (1 x h y h )(1 x 2h y 2h ), ϕ 2 (x, y) = ϕ 3 (x, y) = ϕ 4 (x, y) = ϕ 5 (x, y) = ϕ 6 (x, y) = x ( x 1), 2h h y ( y 1), 2h h x (2 x y ), h h h xy h 2, y (2 x y ). h h h satisfying ϕ i (x j, y j ) = δ ij. The basic functions of Q h in element E are given by ψ 1 (x, y) = 1 x 2h y 2h, ψ 2 (x, y) = ψ 3 (x, y) = x, 2h y. 2h 12 / 21

20 Example The velocity eld: u 1 = x 2 (1 x) 2 2y(1 y)(2y 1), u 2 = y 2 (1 y) 2 2x(1 x)(1 2x). The pressures: p(x, y) = x(1 x)(1 y) The right-hand side: f 1 (x, y) = 4(x 3 (6 12y) + x 4 ( 3 + 6y) + y(1 3y + 2y 2 ) 6xy(1 3y + 2y 2 ) + 3x 2 ( 1 + 4y 6y 2 + 4y 3 )) +y(1 y)(1 2x), f 2 (x, y) = 4( 3( 1 + y) 2 y 2 3x 2 (1 6y + 6y 2 ) +2x 3 (1 6y + 6y 2 ) + x(1 6y + 12y 2 12y 3 + 6y 4 ) +x(1 x)(1 2y)). 13 / 21

21 Results Case: h = h = 1 8 h = 1 18 h = 1 28 L 2 -Errors of u 2.2e e e 03 Iterations of MINRES: TOL=1e / 21

22 Optical ow estimation u t + b u = 0 u(0) = ū p t + b p = 0 p(t ) = (u T u T ) divb = 0 λ b + q = p u b = 0 on Ω 15 / 21

23 Test1 16 / 21

24 Test1 17 / 21

25 Test1 17 / 21

26 Test2 18 / 21

27 Test2 19 / 21

28 Test2 19 / 21

29 Franco Brezzi and Michel Fortin. Mixed and Hybrid Finite Element Methods. Springer-Verlag, / 21

30 Thank you for your attention! 21 / 21