Uniform inf-sup condition for the Brinkman problem in highly heterogeneous media

Transcription

1 Uniform inf-sup condition for the Brinkman problem in highly heterogeneous media Raytcho Lazarov & Aziz Takhirov Texas A&M May 3-4, 2016 R. Lazarov & A.T. (Texas A&M) Brinkman May 3-4, / 30

2 Outline 1 Outline 2 Motivation 3 Related work 4 The continuous problem 5 The discrete problem 6 References R. Lazarov & A.T. (Texas A&M) Brinkman May 3-4, / 30

3 Motivation Brinkman equation ν u + νk 1 u + p = f in Ω, u = 0 in Ω, u = 0 on Ω, where - u is the fluid velocity. - p is the pressure. - ν is the viscosity. We assume ν = 1. - ν is the effective viscosity. We assume ν = 1. - f is the external forcing term. - 0 < K(x) < is the permeability of the medium. R. Lazarov & A.T. (Texas A&M) Brinkman May 3-4, / 30

4 Motivation Brinkman equation Brinkman can be used to model flows in: Pebble Bed Reactors, filtration, biological flows, oil/water reservoirs. R. Lazarov & A.T. (Texas A&M) Brinkman May 3-4, / 30

5 Motivation Flows in highly heterogeneous media Figure: SPE10 3 dimensional permeability distributions, logscale. R. Lazarov & A.T. (Texas A&M) Brinkman May 3-4, / 30

6 Motivation Flows in highly heterogeneous media When permeability field K(x) has large variations and jumps, the problem becomes more challenging. Contrast of the media: κ Ω = max K(x) x Ω min K(x). x Ω Exact solution has low regularity when κ Ω 1. The known iterative methods, converge very slowly or practically do not converge when κ Ω 1, due to the dependence of the condition number of the linear system on κ Ω. R. Lazarov & A.T. (Texas A&M) Brinkman May 3-4, / 30

7 Motivation Preconditioners X - real, separable, Hilbert space, inner product on X is (, ), norm on X is, X be the dual of X,, be the duality pairing. Given A L (X, X ), symmetric and f X, find x X such that Ax = f a(x, y) := Ax, y = f, y y X. R. Lazarov & A.T. (Texas A&M) Brinkman May 3-4, / 30

8 Motivation Preconditioners cont-d Definition [3, Mardal & Winther (2011)] B L (X, X ) is a preconditioner for A L (X, X ) if B is symmetric and positive definite in the sense that is inner product on X. B is a Riesz operator: Given f X, B (Bf, y) = f, y y X. R. Lazarov & A.T. (Texas A&M) Brinkman May 3-4, / 30

9 Motivation Preconditioned system Usually Ax = f is preconditioned as BAx = Bf. Condition number of the system cond (BA) := BA L(X,X ) (BA) 1 L(X,X ) Letting a := sup x,y X a(x, y) x y, inf x X sup y Y cond (BA) a γ. a(x, y) x y γ R. Lazarov & A.T. (Texas A&M) Brinkman May 3-4, / 30

10 Motivation Condition number of the Brinkman problem Let Brinkman system: A = ( ) + IK 1. 0 A ( ) u = p If X = ( H 1 0, ), Q = ( L 2 0, ), then Condition number of the system ( ( ) 1 B = ( ) f I ). cond (BA) O (κ Ω ). R. Lazarov & A.T. (Texas A&M) Brinkman May 3-4, / 30

11 Motivation Question Is it possible to establish well-posedness of the Brinkman problem, such that cond (BA) is independent of the media contrast κ Ω? R. Lazarov & A.T. (Texas A&M) Brinkman May 3-4, / 30

12 Related work Singularly perturbed Stokes system In [2, Mardal & Winther (2004)], authors consider ε u + u + p = f in Ω, u = 0 in Ω, u = 0 on Ω. Authors establish well-posedness in ε-dependent norms, both in continuous and discrete cases. R. Lazarov & A.T. (Texas A&M) Brinkman May 3-4, / 30

13 The continuous problem Intersection and sum of Hilbert spaces If X and Y are Hilbert spaces, then X + Y and X Y are also Hilbert spaces [1, Bergh, Löfström], with the following norms: z X Y = z 2 X + z 2 Y max ( z X, z Y ), z X +Y = If X Y is dense in both X and Y, then inf x 2 z=x+y X + y 2 Y. x X,y Y (X Y ) = X + Y, and (X + Y ) = X Y. R. Lazarov & A.T. (Texas A&M) Brinkman May 3-4, / 30

14 The continuous problem The continuous, weak formulation a(u, v) + b(p, v) = (f, v) b(q, u) = 0, a(u, v) := ( u, v) + (u, v) α, b(p, v) := (p, v), where α = K 1. The natural space for the velocity field is X := ( H 1 0 L 2 α) d, u X := u 2 + u 2 α. R. Lazarov & A.T. (Texas A&M) Brinkman May 3-4, / 30

15 The continuous problem The pressure norm Q should be chosen, so that the Brezzi theory for well posedness holds: a(u, v) a u X v X, a(u, u) a 0 u 2 X, b(p, v) b p Q v X, inf sup b(p,v) p Q p Q v X β. v X For our applications, we need to ensure that cond (BA) is independent of κ Ω. R. Lazarov & A.T. (Texas A&M) Brinkman May 3-4, / 30

16 The continuous problem The pressure norm Let Q := L H 1 K L2 0 = { q L 2 0 : q = q 1 + q 2, q 1 L 2 0, q 2 H 1 K L2 0}, with the associated norm q Q = inf q=q 1 +q 2 q 1 L 2 0,q 2 H 1 K L2 0 q q 2 2 K. R. Lazarov & A.T. (Texas A&M) Brinkman May 3-4, / 30

17 The continuous problem The pressure norm Lemma Given q L 2 0, let q 2 HK 1 L2 0 be the solution of the following elliptic problem: { (K q 2 ) + q 2 = q in Ω, Then K q 2 n = 0 on Ω. q L 2 0 +H q K 1 = L q 2 2 K = q q q 2 2 K = q 2 q 2 2 HK 0. 1 L2 R. Lazarov & A.T. (Texas A&M) Brinkman May 3-4, / 30

18 The continuous problem Well-posedness: Continuity of b(, ) Lemma The bilinear form b(, ) : Q X R is continuous. Proof. q Q, let q = q 1 + q 2 with q 1 L 2 0 and q 2 H 1 K L2 0. Then b (q, v) = b (q 1, v) + b (q 2, v) = (q 1, v) + ( q 2, v) ) = (q 1, v) + (K 1 2 q2, α 1 2 v d q 1 v + q 2 K v α d q q 2 2 K v X. Taking infimum over all q 1, q 2 gives b (q, v) d q Q v X. R. Lazarov & A.T. (Texas A&M) Brinkman May 3-4, / 30

19 The continuous problem Well-posedness: Coercivity of b(, ) Lemma There exists a constant β > 0, independent of 0 < K(x) <, such that inf sup b (q, v) β. q Q v X q Q v X R. Lazarov & A.T. (Texas A&M) Brinkman May 3-4, / 30

20 The continuous problem Proof of the coercivity of b(, ) Proof. The result of Ne cas: q L 2 0 (Ω) : q H 1 (Ω) q. q Q with q = q 1 + q 2 and q i = 0. Assuming that the duality pairing Ω, X X is an extension of the L2 inner product, one obtains that: b (q, v) sup = q X = q v X v H 1 +L 2 X K = inf q q q=q 1 +q 2 H K q 1 H 1, q 2 L 2 K C inf q=q 1 +q 2 q 1 L 2 0,q 2 H 1 K L2 0 = C q Q. q q 2 2 K R. Lazarov & A.T. (Texas A&M) Brinkman May 3-4, / 30

21 The discrete problem Assumption Assumption. The mesh has resolved the heterogeniety of the medium so that elementwise contrast κ E = is a moderate constant. We will also set max K(x) x E min K(x) x E κ Th = max E T h κ E. R. Lazarov & A.T. (Texas A&M) Brinkman May 3-4, / 30

22 The discrete problem Inf-sup with conforming subspaces Fortin s Lemma for Mini-element: Π h = Π b h (I C h) + C h. Π b h satisfies ( Π b h v, q h) = ( v, qh ). C h is Clement or Scott-Zhang (quasilocal) interpolant. In particular need, C h v α c v α, with c independent of κ Ω. For any E T h : C h v 2 α,e max α(x) C hv 2 E C max x E x E α(x) v 2 Ω E max α(x) x E C min α(x) v 2 α,ω E. x Ω E R. Lazarov & A.T. (Texas A&M) Brinkman May 3-4, / 30

23 The discrete problem Non-conforming velocity space X h := J (u h, v h ) : = ( ) Pk d xp k H 0 (div, Ω), Q h := P k 1 H 1 (Ω). e Γ h Γ σ e e e [u h ][v h ], a h (u h, v h ) : = E T h ( u h, v h ) E + (αu h, v h ) + J (u h, v h ) e Γ h Γ e { u h n} [v h ] b h (p h, v h ) : = E T h (p h, v h ) E e Γ h Γ e { v h n} [u h ], R. Lazarov & A.T. (Texas A&M) Brinkman May 3-4, / 30

24 The discrete problem Discrete weak formulation q h Q := a h (u h, v h ) + b h (p h, v h ) = (f, v h ) b h (q h, u h ) = 0. u h Xh := u h 2 E + u h 2 α + J (u h, u h ), E T h inf q q 2 2 q h =q 1 +q K q = h q q 2 2 K. 2 R. Lazarov & A.T. (Texas A&M) Brinkman May 3-4, / 30

25 The discrete problem Discrete inf-sup Lemma The following inf-sup condition holds: There exists a constant β h > 0, independent of κ Ω and h, such that inf q h Q h b h (q h, v h ) sup β h. v h X h q h Q v h Xh R. Lazarov & A.T. (Texas A&M) Brinkman May 3-4, / 30

26 The discrete problem Proof of discrete inf-sup Proof. Let q h Q h. By continuous inf-sup condition, there exists v X such that b (q h, v) β q h Q v X. Let v h = π h v X h be the Raviart-Thomas interpolant of v. By definition of the Raviart-Thomas interpolant: b h (q h, v h ) = ( v h, q h ) E = ( v, q h ) E E T h E T h = b (q h, v) β q h Q v X. One can show that v h α Cκ Th v α v h X C (κ Th ) v X. R. Lazarov & A.T. (Texas A&M) Brinkman May 3-4, / 30

27 The discrete problem Continuous vs. discrete pressure norms Recall that for p h L 2 0, where p h Q = p h 2 p 2 2 H 1 K L2 0, (K p 2, q) + (p 2, q) = (p h, q) q H 1 K L2 0. So in general, p 2 / Q h, and therefore p h Q is not computable. R. Lazarov & A.T. (Texas A&M) Brinkman May 3-4, / 30

28 The discrete problem Continuous vs. discrete pressure norms Let p h Qh = p h 2 p 2,h 2 H 1 K L2 0, where (K p 2,h, q h ) + (p 2,h, q h ) = (p h, q h ) q h Q h, Then p 2,h H 1 K L 2 0 p 2 H 1 K L 2 0 p h Qh p h Q. R. Lazarov & A.T. (Texas A&M) Brinkman May 3-4, / 30

29 The discrete problem The end THANK YOU! R. Lazarov & A.T. (Texas A&M) Brinkman May 3-4, / 30

30 References References J. Bergh, J. Löfström, Interpolation Spaces: An Introduction, Springer Berlin Heidelberg, K-A Mardal, R. Winther, Uniform preconditioners for the time dependent Stokes problem, Numerische Mathematik 98 (2), K-A Mardal, R. Winther, Preconditioning discretizations of systems of partial differential equations, Numer. Linear Algebra Appl. (18) 2011, R. Lazarov & A.T. (Texas A&M) Brinkman May 3-4, / 30