Analysis and Computation of Navier-Stokes equation in bounded domains

Transcription

1 1 Analysis and Computation of Navier-Stokes equation in bounded domains Jian-Guo Liu Department of Physics and Department of Mathematics Duke University with Jie Liu (National University of Singapore) Bob Pego (Carnegie Mellon)

2 2 Outline: The Laplace-Leray commutator and Stokes pressure A formula/decomposition for Navier-Stokes pressure Extended/unconstrained NSE Estimate for the Laplace-Leray commutator Analysis and numerical analysis for (extended) NSE Finite element method with equal order polynomials Stable third order time-split/projection methods Numerics: stability accuracy check, driven cavity, backward-facing step, flow past cylinder

3 3 Navier-Stokes equations for incompressible flow u t + u u + p = ν u + f in Ω momentum u = in Ω incompressibility u = on Ω = Γ no slip Ways to regard and treat pressure: (i) Pressure is like a Lagrange multiplier to enforce incompressibility (ii) Pressure can be eliminated by projection on divergence-free fields or taking curl to get a vorticity equation (iii) Pressure can be found from u and f by solving Poisson equations

4 4 Leray-Helmholtz projection P onto divergence-free fields L 2 (Ω, R N ) = H 1 (Ω) PL 2 (Ω, R N ) v = φ + w, w, ψ = v φ, ψ = for all ψ H 1 (Ω). φ = v, n φ = n v Γ w = in Ω, n w = on Ω. Notation: w = Pv, φ = Qv.

5 5 The Laplace-Leray commutator and Stokes pressure For all v H 2 (Ω, R N ) we have (I P)v = φ = φ = v, Pv = ( )v = v, P v = P( )v ( P P )v = (I P)( )v. We define the Stokes pressure for u H 2 (Ω, R N ) via p S (u) = Q( )u, then p S (u) = ( P P )u. Note ( )u H(div; Ω), hence p S satsifies the BVP p S = in Ω, n p S = n ( )u in H 1/2 (Γ)

6 6 p S arises from tangential vorticity at the boundary Johnston-Liu p S, q = n ( u), q Γ, q H 1 (Ω)

7 Johnston-Liu 7

8 8 Space of Stokes pressures p S = in Ω, n p S = n ( )u = f in H 1/2 (Γ) S p = {p H 1 (Ω)/R : p = in Ω, n p S Γ }, S Γ = {f H 1/2 (Γ) : f = components G of Γ}. G a bounded right inverse p S u from S p H 2 H(Ω, 1 R N ) In R 3, S p is the space of simultaneous gradients and curls: S p = H 1 (Ω) H 1 (Ω, R 3 )

9 9 A formula/decomposition for the Navier-Stokes pressure Suppose that u = Pu is a strong solution of NSE: u t + p = ν u + f u u u = u = in Ω in Ω on Ω Apply P and note P p =, P u = u p S (u): u t + ν p S (u) = ν u + P(f u u) Subtracting, we find that (up to spatial constant) necessarily p = νp S (u) + Q(f u u). p = (f u u), n p = n f n u Γ

10 1 Pressure formula with inflow/outflow Suppose we require: u = g on Ω, where to conserve volume, n g = (t ). Ω Let R(g) solve R(g) = in Ω, n R(g) = n g on Ω. If u = then u R(g) = P(u R(g)) = Pu. Applying P to NSE we now find t (u R(g)) + ν p S (u) = ν u + P(f u u), p = νp S (u) R( t g) + Q(f u u). For q H 1 (Ω) such that: p, q = ν u, n q Γ n g t, q Γ + f u u, q

11 11 Extended/unconstrained NSE u t + u u + p = ν u + f, u Ω = g p = νp S (u) R( t g) + Q(f u u) p = (f u u), n p = n (f u u g t ) n u Γ p, q = ν u, n q Γ n g t, q Γ + f u u, q p, q = ν ( )u, q n g t, q Γ + f u u, q n g t, q Γ + u u + p, q = ν u + f, q ν u, q ( u) t, q + ν ( u), q = ( u) t = ν ( u) in Ω, n ( u) Γ = If ( u) =, then u. Equivalent to the original NSE t=

12 12 Comparing with traditional formulation of NSE u t + P(u u f) = νp u Formally t ( u) = Perform analysis in spaces of divergence-free fields (unconstrained Stokes operator P u is incompletely dissipative). Inf-Sup/LBB condition(ladyzhenskaya-babuška-brezzi) u t + P(u u f) = ν u ν p S = νp u + ν ( u) Improved enforcement of the divergence constraint Formally t ( u) = ν ( u), n ( u) = Equivalent to a reduced formulation of Grubb & Solonnikov (1991) studied as a pseudodifferential IBVP

13 13 Local-time well posedness for extended NSE u t + u u + p = ν u + f, u Ω = p = νp S (u) R( t g) + Q(f u u) Assume u in H uin := H 1 (Ω, R N ), f H f := L 2 (, T ; L 2 (Ω, R N )), g H g := H 3/4 (, T ; L 2 (Γ, R N )) L 2 (, T ; H 3/2 (Γ, R N )) {g t (n g) L 2 (, T ; H 1/2 (Γ))}, g = u in (t =, x Γ), n t g, 1 Γ = Then T > so that a unique strong solution exists, with u L 2 ([, T ], H 2 ) H 1 ([, T ], L 2 ) C([, T ], H 1 ).

14 14 Estimate for the Laplace-Leray commutator Recall [, P]u = ( P P )u = p S For period box [P, ]u = If u =, then [, P]u = ((I P) )u u. Theorem Let Ω R N (N 2) be a bounded C 3 domain Then ε > C such that for all u H 2 H(Ω, 1 R N ), ( ) 1 ( P P )u ε u 2 + C u 2 Ω Hence our unconstrained NSE is fully dissipative: u t + P(u u f) + ν p S = ν u NSE as perturbed heat equation! Ω Ω

15 15 An example in half period strip < x < 2π, y u(x, y) = sin(kx)ye ky, any v(x, y) with v(x, ) = Stokes pressure: p(x, y) = cos(kx)e ky p =, y p(x, ) = k cos(kx) = x ( y u x v) y= u = 2k sin(kx)e ky = 2 x p, (u, ) =, (, v) = at y = (boundary properties ) p x 2 = p y 2 = πk 2 (equal partition) u x p y p, x p = p x 2 p y 2 = (orthogonality) y p This implies: p 2 = 1 2 u ( u 2 + v 2 )

16 16 Basic ingredients in the (elementary) proof: Let Φ(x) = dist(x, Γ), n(x) = Φ(x) Ω s = {x Ω Φ(x) < s}, f, g s := f, g Ωs Choose a cutoff function ξ so ξ(x) = 1 (Φ < 1 2s), ξ(x) = (Φ > s). Decompose velocity field into parts parallel and normal to Γ: u = u + u, u = ξnn t u + (1 ξ)u, u = ξ(i nn t )u. (i) A parallel-normal decomposition u = u + u satisfying on Ω: u =, u = (ii) An orthogonality identity: u 2 = p S 2 + p S u 2 (iii) Sharp N2D estimates on tubes: p S 2 s p S 2 s

17 17 p S = ( P P )u = (I P)( )(u + u ) (due to u = on Ω) p S = u u P( )u p S, p S = p S, u p S, u p S, Pv = p S, u (due to u = on Ω, p S =, Pv) u 2 = p S 2 + p S u 2 (orthogonality)

18 18 p u 2 = p 2 Ω c + ( p u s ) 2 s + ( p u ) 2 s p 2 Ω c + (1 ε) p 2 s s + ( p u ) 2 s + = ( p u ), p s + ( p u ), p s + Hence Φ>s p 2 2 ( p u ), p s 2 ( p u ), p s ( p u ) 2 s + p 2 s ( p u ) 2 s p 2 s + 2( p 2 s p 2 s) junk p u 2 (1 ε) p 2 + 2( p 2 s p 2 s) junk

19 19 D2N/N2D bounds on tubes Ω s = {x Ω Φ(x) < s} Lemma For s > small C such that whenever p = in Ω s and < s < s then n p 2 (I nn t ) p 2 C s p 2 Φ<s Φ<s In the limit s, it reduce to n p 2 (I nn t ) p 2 C Γ Γ Ω p 2 In a half space: n p 2 = (I nn t ) p 2. Known as Rellich identity in 2D circular disk.

20 2 u 2 = u u, u + u 2 (1 ε) u 2 C u 2 = (1 ε)( p S 2 + p S u 2 ) C u 2 (2 3ε) p S 2 C u 2

21 21 Planar domains with corners a negative result Theorem (Cozzi & Pego, 21) Let Ω R 2 be a bounded domain with a locally straight corner. Given any β < 1 and C >, there exists u H 2 H 1 (Ω, R N ) such that Ω p S (u) 2 > β Ω u 2 + C Ω u 2 Open Q: Is extended Navier-Stokes dynamics well-posed in Lipschitz (or polygonal) domains?

22 22 Difference scheme with pressure explicit (Johnston-L, 24) u n+1 u n t p n = (f n u n u n ) in Ω n p n = n f n νn ( u n ) on Ω ν u n+1 = f n u n u n p n u n+1 = on Ω No Stokes solver! u n is not forced to be zero Demonstrated numerical stability, full time accuracy Proved linear stability of tangential velocity in slab geometry by energy estimates Equivalent to gauge method (E & L). Unconditional stability for schemes with time lag BC for Stokes eq (Wang & L, 2)

23 23 Time-discrete scheme with pressure explicit (related to projection methods: Ti96,Pe1,GS3,JL4) u n+1 u n t ν u n+1 = f n u n u n p n, u n+1 = on Ω, p n = ν p n + (I P)(f n u n u n ). S We have the pressure estimate, for fixed constant β < 1 p n ν p n + f n u n u n S νβ u n + f n u n u n + C u n

24 24 Stability analysis in H 1 : dot with u n+1 u n+1 u n t ν u n+1 = f n u n u n p n u n+1 u n 2 + u n+1 2 u n 2 + ν u n t u n+1 (2 f n u n u n + ν p n ) S ( ε ν 2 ) un ε 1 f n u n u n 2 + ν 2 p S 2 This gives u n+1 2 u n 2 t + (ν ε 1 ) u n ε 1 ( f n 2 + u n u n 2 ) + ν p n S 2

25 25 Handling the pressure u n+1 2 u n 2 + (ν ε 1 ) u n+1 2 t 8 ( f n 2 + u n u n 2 ) + ν p n 2 S ε 1 Use the theorem (β = ε): ν pn S 2 νβ u n 2 + νc u n 2 u n+1 2 u n 2 t + (ν ε 1 )( u n+1 2 u n 2 ) + (ν ε 1 νβ) u n 2 8 ε 1 ( f n 2 + u n u n 2 ) + νc u n 2

26 26 Handling the nonlinear term Use Ladyzhenskaya s inequalities to get for N = 2, 3 (standard) u n u n 2 ε 2 u n 2 + 4C ε 2 u n 6 Take ε 1, ε 2 small so ε := (ν ε 1 νβ ε 2 ) > to get: u n+1 2 u n 2 + ε u n 2 C( f n 2 + u n 2 + u n 6 ) An easy discrete Gronwall inequality leads to:

27 27 Stability theorem for N = 2, 3 Theorem Take f L 2 (, T ; L 2 (Ω, R N )), u H 2 H 1 (Ω, R N ). Then positive constants T and C depending only upon Ω, ν and M := u 2 + ν t u 2 + T f 2, so that whenever n t T we have sup u k 2 + k n n 1 k= ( u k 2 + u k+1 u k 2) t C. t

28 28 Classic inf-sup condition for steady Stokes flow FEM Find u X H(Ω, 1 R N ), p Y L 2 (Ω)/R so that u, v + p, v = f, v v X, u, q = q Y. Well-known: inf sup p Y v X p, v p v c >. (1) Condition (1) is necessary and sufficient for stability of (p, u) in L 2 H 1 (the weak NS solution space). Many simple finite elements fail this condition (e.g., equal-order Lagrange)

29 29 Steady Stokes flow in terms of Stokes pressure Find u X H 2 H(Ω, 1 R N ), p Y H 1 (Ω)/R so that u, v + p, v = f, v v X, p, q u, n q Γ = f, q q Y. Stability of (p, u) H 1 (H 2 H 1 (Ω, R N )) (strong solution space) is not subject to standard inf-sup velocity-pressure compatibility. Still a mixed formulation computationally for steady flow Possible gain in simplicity of discretization with C elements (e.g. for complex fluid-coupled systems) nonconforming for H 2

30 3 C Finite element schemes (Johnston-Liu, 4) C finite element scheme: X h H(Ω, 1 R N ), all v h X h and q h Y h, we require require Y h H 1 (Ω)/R. For p n h, q h = f n u n h u n h, q h + ν u n h, n q h Γ, un+1 h u n h t, v h + ν u n+1 h, v h = f n u n h u n h p n h, v h. This scheme has fully accuracy for both velocity and pressure when there are enough resolution and regularity for the solution. We have no theorem on why it works. For C 1 FEM without inf-sup compatibility, however, we have fully theory: stability and error estimates (both for velocity and pressure).

31 31 Error estimates of C 1 FE scheme u n h X h H 2 H 1 (Ω, R N ), p n h Y h H 1 (Ω)/R. For all v h X h and q h Y h, require 1 t ( u n+1 h, v h u n h, v h ) + ν u n+1 h, v h = p n h, v h f n u n h u n h, v h. p n h, q h = f n u n h u n h, q h + ν u n h, n q h Γ, Theorem Assume Ω is a bounded domain in R N (N=2,3) with C 3 boundary. Let M, >, and let T > be given by the stability theorem. Let m 2, m 1 be integers, and assume For any v H m+1 H 1 (Ω, R N ) and q H m (Ω), inf v h X,h (v v h ) C h k 1 v H k+1 for 2 k m,

32 32 inf (q q h ) C h m 1 q q h Y H m, h where C > is independent of v, q and h. Then there exists C 1 > with the following property. Whenever u h X h, < h < 1, < n t min(t, T ), and n u h 2 + t u h 2 + f(t k ) 2 t M, then e n = u(t n ) u n h, rn = p(t n ) p n h of C1 finite element scheme satisfy sup e k 2 + k n k= n ( e k 2 + r k 2) t k= C 1 ( t 2 + h 2m 2 + h 2m 2 + e 2 + e 2 t).

33 33 3. Split-step time discretizations of NSE Chorin/Temam projection method Given velocity U n = PU n, let F n+1 = f n+1 U n U n and update velocity by solving u n+1 U n = ν u n+1 + F n+1 t in Ω, u n+1 = on Ω, U n+1 = u n+1 φ n+1 = Pu n+1 Why does it work? Apply P. Stokes pressure reappears implicitly: U n+1 U n + ν p S (U n+1 ) = ν U n+1 + PF n+1, t U n+1 = φ n+1 on Ω (slip error)

34 34 Subtract Formally we get a consistent 1st-order scheme, since φ n+1 t ν φ n+1 = νp S (U n+1 ) + Q(f n+1 U n U n ), n φ n+1 = on Ω. thus φ n+1 Ω = O( t). The correct corresponding pressure p n+1 = νp S (U n+1 ) + Q(f n+1 U n+1 U n+1 ) = φn+1 t ν φ n+1 + O( t) The (old) pressure approximation φ n+1 / t has a boundary layer with O(1) gradient error on Ω, since n φ n+1 = but n p n+1 = n (ν U n+1 + f n+1 U n+1 U n+1 ).

35 35 Higher-order accurate time discretization Chorin-Temam method: 1st-order in time, boundary layers for p Brown, Cortez, Minion 1: Explain 2nd-order time accuracy Guermond, Minev, Shen 6: Review many projection methods, FEM use higher-order backward time difference formulas, and improve boundary-layer accuracy using various strategies, e.g.: Pressure approximation (Karniadakis et al 91): extrapolate curl-curl BCs to approximate p n+1 in a RHS F n+1 Pressure update (van Kan 86, BCG 89): predict then correct pressure, avoiding curl-curl BCs Slip correction (Kim-Moin 86): extrapolate to improve BCs for U n+1

36 36 A stable (3,2) slip-corrected projection method Use BD3 formula D 3 u n+1 = 1 t ( α u n+1 + α k u n+1 k) k 1 Write H j = U j U j & extrapolate: E 3 H n+1 = 3H n 3H n 1 + H n 2, E 2 φ n+1 = 2φ n φ n Given U j u(j t) (j n), solve (without any pressure!) 1 ( α u n+1 t + k 1 α k U n+1 k) = ν u n+1 + f n+1 E 3 H n+1, u n+1 = g n+1 + E 2 φ n+1 on Γ. 2. U n+1 = u n+1 φ n+1, φ n+1 = u n+1, n φ n+1 =.

37 37 Formal accuracy: Apply P and recall Pu j = PU j = U j R(g j ), P u n+1 = Pu n+1 p S (u n+1 ) = U n+1 p S (U n+1 ) : D 3 U n+1 + p n+1 = ν U n+1 + f n+1 E 3 H n+1, p n+1 = νp S (U n+1 ) R(D 3 g n+1 ) + Qf n+1 QE 3 H n+1, U n+1 = g n+1 (φ n+1 E 2 φ n+1 ) on Γ. ( α ) t ν (φ n+1 E 2 φ n+1 ) = p n+1 2 p n + p n 1 = O( t 2 ), hence U n+1 g n+1 = O( t 3 ) formally and the scheme is formally 3rd-order accurate overall.

38 38 A stable (3,2) pressure approximation method Update intermediate velocity u n via (H n = u n u n ) F = f n+1 E 3 H n+1 1 t j 1 α j u n+1 j, P = νp S (E 2 u n+1 ) α t R(gn+1 ) + QF α t un+1 ν u n+1 = P + F u n+1 = g n+1 on Γ The div-free velocity and consistent pressure are U n+1 = u n+1 φ n+1, φ n+1 = u n+1, n φ n+1 =, p n+1 = P ν u n+1. (Improves over previous methods)

39 39 (3,3) pressure update method (cf. BCG 89, Ren et al 5) 1. Find u n+1 to solve 1 ( α u n+1 t + k 1 α k U n+1 k) + E 3 p n+1 = ν u n+1 + f n+1 E 3 H n+1, u n+1 = g n+1 on Γ. 2. U n+1 = u n+1 φ n+1, φ n+1 = u n+1, n φ n+1 =. 3. Update pressure via p n+1 = E 3 p n+1 + ( α t ν ) φ n+1

40 4 Single mode Stokes test problem in strip 1 < x < 1 t u + p S (u) = u, u x=±1 = u(t, x, y) = e iξy σt (u(x, ξ), iv(x, ξ)) For the case of odd symmetry: p = e iξy σt sinh ξx, ( ) cosh ξx cos µx u(x) = A cosh ξ cos µ ( ) sinh ξx sin µx v(x) = B sinh ξ sin µ, A = ξ cosh ξ ξ 2 + µ 2,, B = ξ sinh ξ ξ 2 + µ 2. ξ tanh ξ + µ tan µ =, σ = ξ 2 + µ 2. Usually take ξ = 1, µ , σ

41 41 Single-mode time-discrete stability tests A u n+1 + A 1 u n A k u n+1 k =. Look for u n = κ n u with I I A 3 A 2 A 1 u κu κ 2 u = κ I I A u κu κ 2 u. Solve for eigenvalues κ using Matlab eigs. Plot max κ vs t for various combinations of finite-element velocity/pressure pairs and orders of accuracy.

42 42 Slip-correction schemes, single-mode stability test (3,2) P1/P1 (3,2) P2/P1 (2,2) P1/P1 (2,2) P2/P1 (4,3) P1/P1 (4,3) P2/P t 1: max κ vs t. 3 elements for each var. Solid lines: space-continuous theory. (2,2), (3,2): unconditional stability. (cf. Leriche et al 6 w/pa) (3,3), (4,3): stability for t < t c independent of h, ξ. (smooth?)

43 43 Pressure update schemes, single-mode stability test (3,2) P1/P1 (3,2) P2/P1 (2,2) P1/P1 (2,2) P2/P1 (3,3) P1/P1 (3,3) P2/P1 (4,3) P1/P1 (4,3) P2/P t 2: max κ vs t. 3 elements for each var. Solid lines: interpolated. All P1/P1 PU schemes: unstable. P2/P1 velocity-pressure pairs: (2,2), (3,2): unconditional stability. (3,3),(4,3): stability window

44 44 2D test problem: stability & accuracy Ω = square [ 1, 1] 2 hole, or ring. ν = 2, t = 2 u(t) = g(t) ( ) cos 2 (πx/2) sin(πy) sin(πx) cos 2, (πy/2) p(t) = g(t) cos(πx/2) sin(πy/2).

45 45 2D test: stability in domains with & without corners t c \ N 1 2 (3,3) P1/P (3,3) P2/P (4,3) P1/P (4,3) P2/P Largest time step for linear stability of PA schemes in a square with hole. N = # of refinements from grid in figure. Diffusive constraint on t. t c \ N 1 2 (3,3) P1/P (3,3) P2/P (4,3) P1/P (4,3) P2/P Largest time step for linear stability of PA schemes in annulus. N = # of refinements from grid in figure. No diffusive constraint on t.

46 46 Accuracy: slip-corrected P4 FEM, square with hole log 1 E (& local order α) vs. t E \ t p ˆp h L (2.71) 5.8 (2.98) 5.99 (3.2) (p ˆp h ) L (3.73) 4.39 (3.4) 5.17 (2.62) u u h L 5.75 (2.73) 6.65 (3.2) 7.57 (3.5) (u u h ) L 3.94 (3.22) 4.95 (3.35) 5.83 (2.92) Top: (3,2) SC scheme. Bottom: (2,1) SC scheme. E \ t p ˆp h L (1.97) 2.64 (2.2) 3.26 (2.4) (p ˆp h ) L (1.96) 2.15 (2) 2.76 (2.2) u u h L 3.61 (1.96) 4.22 (2.2) 4.84 (2.5) (u u h ) L 2.19 (1.97) 2.8 (2.1) 3.41 (2.4)

47 47 Pressure error for square with hole, (3,2) schemes x 1 5 x Pressure error for the (3,2) PA (left) and SC (right) schemes. t = P4 elements for each variable. Only values at vertices of triangles are used in the plots.

48 48 Benchmark tests with equal-order C finite elements (u,v)=(1,) (u,v) given (u,v) given 3: Mesh used for backward facing step flow when ν = 1/6 and for flow past a cylinder when ν = 1/1.

49 .5 49 Driven Cavity, Re=1, t= stretched rectangular grid 8192 piecewise linear C elements for each variable. h min =.594, t =.6, (3,2) slip-corrected scheme. Left: vorticity contours. Right: pressure contours.

50 5 Backward-facing step.5 S X (3,2) SC scheme, 664 P1 elements for each variable (dof=3487). ν = 1/1, t = 2, h min =.783, t =.6..5 X 1 X 2 S X (3,2) SC scheme, 17 P2 elements for each variable (dof=3925). ν = 1/6, t = 12, h min =.186, t =.3.

51 51 Flow past cylinder, ν = 1/

52 52 Drag, lift, pressure drop t(c d,max )=3.9348, c d,max = t t(c l,max )=5.6932, c l,max = t p max =2.3219, p(8)= t (3,2) SC scheme 763 isoparametric P4 elements (dof=6322) for each variable.

53 53 Summary The Navier-Stokes pressure for strong solutions in C 3 domains is determined by the current velocity and data by p = νp S (u) R( t g) + Q(f u u). The Laplace-Leray commutator is p S (u) and is strictly controlled by u at leading order, when u = on the boundary. Improved accuracy (esp. for pressure) in numerical computations Improved flexibility of finite-element methods inf-sup conditions need not be respected. Short-time wellposedness extends to u with simple proof

54 Thank you! 54