Analysis and Computation of Navier-Stokes Equation in Bounded Domains

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 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 Navier-Stokes equations for incompressible ﬂow

ut + u · ∇u + ∇p = ν∆u + f in Ω momentum ∇ · u = 0 in Ω incompressibility u = 0 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 Leray-Helmholtz projection P onto divergence-free ﬁelds

L2(Ω, RN ) = ∇H1(Ω) ⊕ PL2(Ω, RN ) v = ∇φ + w,

hw, ∇ψi = hv − ∇φ, ∇ψi = 0 for all ψ ∈ H1(Ω).

∆φ = ∇ · v, ∂nφ = n · v Γ ∇ · w = 0 in Ω, n · w = 0 on ∂Ω.

Notation: w = Pv, φ = Qv. 5 The Laplace-Leray commutator and Stokes pressure

For all v ∈ H2(Ω, RN ) 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 ∈ H2(Ω, RN ) via

pS(u) = Q(∆ − ∇∇·)u, then ∇pS(u) = (∆P − P∆)u.

• Note (∆ − ∇∇·)u ∈ H(div; Ω), hence pS satsiﬁes the BVP

−1/2 ∆pS = 0 in Ω, n · ∇pS = n · (∆ − ∇∇·)u in H (Γ) 6

pS arises from tangential vorticity at the boundary

Johnston-Liu

1 h∇pS, ∇qi = −hn × (∇ × u), ∇qiΓ, ∀q ∈ H (Ω) 7

Johnston-Liu 8 Space of Stokes pressures

−1/2 ∆pS = 0 in Ω, n · ∇pS = n · (∆ − ∇∇·)u = f in H (Γ)

1 Sp = {p ∈ H (Ω)/R : ∆p = 0 in Ω, n ·∇p ∈ SΓ}, Z −1/2 SΓ = {f ∈ H (Γ) : f = 0 ∀components G of Γ}. G

2 1 N •∃ a bounded right inverse ∇pS 7→ u from Sp → H ∩ H0 (Ω, R ) 3 • In R , ∇Sp is the space of simultaneous gradients and curls:

1 1 3 ∇Sp = ∇H (Ω) ∩ ∇ × H (Ω, R ) 9 A formula/decomposition for the Navier-Stokes pressure Suppose that u = Pu is a strong solution of NSE:

ut + ∇p = ν∆u + f − u · ∇u in Ω ∇ · u = 0 in Ω u = 0 on ∂Ω

Apply P and note P∇p = 0, P∆u = ∆u − ∇pS(u):

ut + ν∇pS(u) = ν∆u + P(f − u · ∇u)

Subtracting, we ﬁnd that (up to spatial constant) necessarily

p = νpS(u) + Q(f − u · ∇u).

∆p = ∇ · (f − u · ∇u), ∂np = n · f − n · ∇ × ∇ × u Γ 10 Pressure formula with inﬂow/outﬂow

Suppose we require: u = g on ∂Ω, Z where to conserve volume, n · g = 0 (t ≥ 0). ∂Ω Let R(g) solve ∆R(g) = 0 in Ω, n · ∇R(g) = n · g on ∂Ω. If ∇ · u = 0 then u − ∇R(g)= P(u − ∇R(g)) = Pu. Applying P to NSE we now ﬁnd

∂t(u − ∇R(g)) + ν∇pS(u)= ν∆u + P(f − u · ∇u),

p = νpS(u) −R (∂tg)+ Q(f − u · ∇u). For ∀q ∈ H1(Ω) such that:

h∇p, ∇qi = νh∇ × u, n×∇qiΓ − hn·gt, qiΓ + hf − u · ∇u, ∇qi 11 Extended/unconstrained NSE

ut + u · ∇u + ∇p = ν∆u + f, u|∂Ω = g

p = νpS(u) −R (∂tg)+ Q(f − u · ∇u)

∆p = ∇·(f −u·∇u), ∂np = n·(f −u·∇u−gt)−n·∇×∇×u Γ h∇p, ∇qi = νh∇ × u, n×∇qiΓ − hn·gt, qiΓ + hf − u · ∇u, ∇qi

h∇p, ∇qi = νh(∆ − ∇∇·)u, ∇qi − hn·gt, qiΓ + hf − u · ∇u, ∇qi hn·gt, qiΓ + hu · ∇u + ∇p, ∇qi = hν∆u + f, ∇qi − νh∇∇ · u, ∇qi h(∇ · u)t, qi + νh∇(∇ · u), ∇qi = 0

(∇ · u)t = ν∆(∇ · u) in Ω, ∂n(∇ · u)|Γ = 0

If (∇ · u) = 0, then ∇ · u ≡ 0. Equivalent to the original NSE t=0 12 Comparing with traditional formulation of NSE

ut + P(u · ∇u − f) = νP∆u

• Formally ∂t(∇ · u) = 0 • Perform analysis in spaces of divergence-free ﬁelds (unconstrained Stokes operator P∆u is incompletely dissipative).

• Inf-Sup/LBB condition(Ladyzhenskaya-Babuˇska-Brezzi)

ut + P(u · ∇u − f) = ν∆u − ν∇pS = νP∆u + ν∇(∇ · u) • Improved enforcement of the divergence constraint Formally ∂t(∇ · u) = ν∆(∇ · u), ∂n(∇ · u) = 0 • Equivalent to a ‘reduced’ formulation of Grubb & Solonnikov (1991) studied as a pseudodiﬀerential IBVP 13 Local-time well posedness for extended NSE

ut + u · ∇u + ∇p = ν∆u + f, u|∂Ω = 0

p = νpS(u) −R (∂tg)+ Q(f − u · ∇u)

Assume 1 N uin ∈ Huin := H (Ω, R ), 2 2 N f ∈ Hf := L (0,T ; L (Ω, R )), 3/4 2 N 2 3/2 N g ∈ Hg := H (0,T ; L (Γ, R )) ∩ L (0,T ; H (Γ, R )) 2 −1/2 ∩{g | ∂t(n · g) ∈ L (0,T ; H (Γ))},

g = uin (t = 0, x ∈ Γ), hn · ∂tg, 1iΓ = 0 Then ∃T ∗ > 0 so that a unique strong solution exists, with

u ∈ L2([0,T ∗],H2) ∩ H1([0,T ∗],L2) ,→ C([0,T ∗],H1). 14 Estimate for the Laplace-Leray commutator

Recall [∆, P]u = (∆P − P∆)u = ∇pS For period box [P, ∆]u = 0 If ∇ · u = 0, then k[∆, P]uk = k((I − P)∆ − ∇∇·)uk ≤ k∆uk.

Theorem Let Ω ⊂ RN (N ≥ 2) be a bounded C3 domain Then 2 1 N ∀ε > 0 ∃C ≥ 0 such that for all u ∈ H ∩ H0 (Ω, R ), Z 1 Z Z |(∆P − P∆)u|2 ≤ + ε |∆u|2 + C |∇u|2 Ω 2 Ω Ω

Hence our unconstrained NSE is fully dissipative:

ut + P(u · ∇u − f)+ ν∇pS = ν∆u

NSE as perturbed heat equation! 15 An example in half period strip 0 < x < 2π, y ≥ 0

u(x, y) = sin(kx)ye−ky, any v(x, y) with v(x, 0) = 0

Stokes pressure: p(x, y) = cos(kx)e−ky

∆p = 0, ∂yp(x, 0) = −k cos(kx) = −∂x(∂yu − ∂xv)|y=0 −ky ∆u = −2k sin(kx)e = 2∂xp, ∇ · (u, 0) = 0, ∇ × (0, v) = 0 at y = 0 (boundary properties )

2 2 πk kpxk = kpyk = 2 (equal partition) ∆u − ∂xp ∂xp 2 2 , = kpxk − kpyk = 0 (orthogonality) 0 − ∂yp ∂yp

1 1 This implies: k∇pk2 = k∆uk2 ≤ (k∆uk2 + k∆vk2) 2 2 16 Basic ingredients in the (elementary) proof:

Let Φ(x) = dist(x, Γ), n(x) = −∇Φ(x)

Ωs = {x ∈ Ω | Φ(x) < s}, hf, gis := hf, giΩs 1 Choose a cutoﬀ function ξ so ξ(x) = 1 (Φ < 2s), ξ(x) = 0 (Φ > s). Decompose velocity ﬁeld into parts parallel and normal to Γ:

t t u = u⊥ + uk, u⊥ = ξnn u + (1 − ξ)u, uk = ξ(I − nn )u.

(i) A parallel-normal decomposition u = uk + u⊥

satisfying on ∂Ω: ∇ · uk = 0, ∇ × u⊥ = 0 2 2 2 (ii) An orthogonality identity: k∆ukk = k∇pSk + k∇pS − ∆ukk 2 2 (iii) Sharp N2D estimates on tubes: k∇pSkks ∼ k∇pS⊥ks 17

H¨¨H ∇pS = (∆P − P∆)u = (I − P)(∆ − ∇∇·)(uk + ¨u¨⊥HH)

(due to ∇ × u⊥ = 0 on ∂Ω)

∇pS = ∆uk − ∇∇ · uk − P(∆ − ∇∇·)uk

h∇pS, ∇pSi = h∇pS, ∆uki − h∇pS, ∇∇ · uki − h∇pS, Pvi

= h∇pS, ∆uki − 0 − 0

(due to ∇ · uk = 0 on ∂Ω, ∆pS = 0, ∇ ⊥ Pv)

2 2 2 k∆ukk = k∇pSk + k∇pS − ∆ukk (orthogonality) 18

2 2 2 2 k∇p − ∆u k = k∇pk c + k(∇p − ∆u ) k + k(∇p − ∆u ) k k Ωs k ⊥ s k k s 2 2 2 ≥ k∇pk c + (1 − ε)k∇p k + k(∇p − ∆u ) k + ··· Ωs ⊥ s k k s Z 2 0 = h(∇p − ∆uk)k, ∇pkis + h(∇p − ∆uk)⊥, ∇p⊥is + |∇p| Φ>s Hence

2h(∇p − ∆uk)⊥, ∇p⊥is ≤ −2h(∇p − ∆uk)k, ∇pkis 2 2 ≤k (∇p − ∆uk)kks + k∇pkks

2 2 2 2 k(∇p − ∆uk)kks ≥ k∇pkks + 2(k∇p⊥ks − k∇pkks) − junk 2 2 2 2 k∇p − ∆ukk ≥ (1 − ε)k∇pk + 2(k∇p⊥ks − k∇pkks) − junk 19

D2N/N2D bounds on tubes Ωs = {x ∈ Ω | Φ(x) < s}

Lemma For s0 > 0 small ∃C0 such that whenever ∆p = 0 in Ωs0 and 0 < s < s0 then Z Z 2 t 2 2 |n·∇p| − |(I − nn )∇p| ≤ C0s |∇p| Φ

~~In the limit s → 0, it reduce to Z Z Z 2 t 2 2 |n·∇p| − |(I − nn )∇p| ≤ C0 |∇p| Γ Γ Ω~~

~~In a half space: kn·∇pk2 = k(I − nnt)∇pk2. Known as Rellich identity in 2D circular disk. 20~~

~~2 2 2 k∆uk = k∆ukk + 2h∆uk, ∆u⊥i + k∆u⊥k~~

~~2 2 ≥ (1 − ε)k∆ukk − Ck∇uk~~

~~2 2 2 = (1 − ε)(k∇pSk + k∇pS − ∆ukk ) − Ck∇uk~~

~~2 2 ≥ (2 − 3ε)k∇pSk − Ck∇uk 21 Planar domains with corners – a negative result~~

Theorem (Cozzi & Pego, 2010) Let Ω ⊂ R2 be a bounded domain with a locally straight corner. Given any β < 1 and C > 0, there 2 1 N exists u ∈ H ∩ H0 (Ω, R ) such that Z Z Z 2 2 2 |∇pS(u)| > β |∆u| + C |∇u| Ω Ω Ω

~~Open Q: Is extended Navier-Stokes dynamics well-posed in Lipschitz (or polygonal) domains? 22 Diﬀerence scheme with pressure explicit (Johnston-L, 2004)~~

~~∆pn = ∇ · (f n − un · ∇un) in Ω n ·∇pn = n · f n − νn · (∇ × ∇ × un) on ∂Ω un+1 − un − ν∆un+1 = f n − un · ∇un − ∇pn ∆t un+1 = 0 on ∂Ω~~

• No Stokes solver! ∇ · un 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, 2000) 23 Time-discrete scheme with pressure explicit (related to projection methods: Ti96,Pe01,GS03,JL04)

~~un+1 − un − ν∆un+1 = f n − un · ∇un − ∇pn, ∆t un+1 = 0 on ∂Ω,~~

~~n n n n n ∇p = ν∇pS + (I − P)(f − u · ∇u ). We have the pressure estimate, for ﬁxed constant β< 1~~

~~n n n n n k∇p k ≤ νk∇pS k + kf − u · ∇u k~~

~~≤ νβk∆unk + kf n − un · ∇unk + Ck∇unk 24~~

~~1 n+1 Stability analysis in H0 : dot with −∆u~~

~~un+1 − un − ν∆un+1 = f n − un · ∇un − ∇pn ∆t~~

~~k∇un+1 − ∇unk2 + k∇un+1k2 − k∇unk2 + νk∆un+1k2 2∆t n+1 n n n n ≤ k∆u k(2kf − u · ∇u k + νk∇pS k)~~

ε1 ν n+1 2 2 n n n 2 ν 2 ≤ ( + )k∆u k + kf − u · ∇u k + k∇pSk 2 2 ε1 2 This gives k∇un+1k2 − k∇unk2 + (ν − ε )k∆un+1k2 ∆t 1 8 n 2 n n 2 n 2 ≤ (kf k + ku · ∇u k ) + νk∇pS k ε1 25 Handling the pressure

~~k∇un+1k2 − k∇unk2 + (ν − ε )k∆un+1k2 ∆t 1 8 n 2 n n 2 n 2 ≤ (kf k + ku · ∇u k ) + νk∇pS k ε1~~

~~1 n 2 n 2 n 2 Use the theorem (β = 2 + ε): νk∇pS k ≤ νβk∆u k + νCk∇u k~~

~~k∇un+1k2 − k∇unk2 + (ν − ε )(k∆un+1k2 − k∆unk2) ∆t 1 n 2 + (ν − ε1 − νβ)k∆u k 8 ≤ (kf nk2 + kun · ∇unk2) + νCk∇unk2 ε1 26 Handling the nonlinear term~~

~~Use Ladyzhenskaya’s inequalities to get for N = 2, 3 (standard)~~

~~n n 2 n 2 4C n 6 ku · ∇u k ≤ ε2k∆u k + k∇u k ε2~~

~~Take ε1, ε2 small so ε := (ν − ε1 − νβ − ε2) > 0 to get:~~

~~k∇un+1k2 − k∇unk2 + εk∆unk2~~

~~≤ C(kf nk2 + k∇unk2 + k∇unk6)~~

~~An easy discrete Gronwall inequality leads to: 27 Stability theorem for N = 2, 3~~

~~2 2 N 0 2 1 N Theorem Take f ∈ L (0,T ; L (Ω, R )), u ∈ H ∩ H0 (Ω, R ). Then ∃ positive constants T ∗ and C∗ depending only upon Ω, ν and~~

~~Z T 0 2 0 2 2 M0 := k∇u k + ν∆tk∆u k + kfk , 0 so that whenever n∆t ≤ T ∗ we have~~

~~n−1 k+1 k 2! k 2 X k 2 u − u ∗ sup k∇u k + k∆u k + ∆t ≤ C . 0≤k≤n ∆t k=0 28 • Classic inf-sup condition for steady Stokes ﬂow FEM~~

~~1 N 2 Find u ∈ X ⊂ H0 (Ω, R ), p ∈ Y ⊂ L (Ω)/R so that~~

~~h∇u, ∇vi + h∇p, vi = hf, vi ∀v ∈ X, hu, ∇qi = 0 ∀q ∈ Y. hp, ∇ · vi inf sup ≥ c > 0. (1) p∈Y v∈X kpkk∇vk Well-known:~~

• Condition (1) is necessary and sufficient for stability of 2 1 (p, u) in L × H0 (the weak NS solution space). • Many simple finite elements fail this condition (e.g., equal-order Lagrange) 29 Steady Stokes flow in terms of Stokes pressure

~~2 1 N 1 Find u ∈ X ⊂ H ∩ H0 (Ω, R ), p ∈ Y ⊂ H (Ω)/R so that~~

~~h∇u, ∇vi + h∇p, vi = hf, vi ∀v ∈ X,~~

~~h∇p, ∇qi − h∇ × u, n × ∇qiΓ = hf, ∇qi ∀q ∈ Y.~~

~~1 2 1 N Stability of (p, u) ∈ H × (H ∩ H0 (Ω, R )) (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 C0 elements (e.g. for complex ﬂuid-coupled systems) — nonconforming for H2 30~~

~~C0 Finite element schemes (Johnston-Liu, ’04)~~

~~0 1 N 1 C ﬁnite element scheme: Xh ⊂ H0 (Ω, R ),Yh ⊂ H (Ω)/R. For all vh ∈ Xh and qh ∈ Yh, we require require~~

~~n n n n n h∇ph, ∇qhi = hf − uh ·∇uh, ∇qhi + νh∇ × uh, n×∇qhiΓ,~~

un+1 − un h h h , v i + νh∇un+1, ∇v i = hf n − un · ∇un − ∇pn, v i. ∆t h h h h h h 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 C1 FEM without inf-sup compatibility, however, we have fully theory: stability and error estimates (both for velocity and pressure). 31~~

~~Error estimates of C1 FE scheme~~

~~n 2 1 N n 1 uh ∈ Xh ⊂ H ∩ H0 (Ω, R ), ph ∈ Yh ⊂ H (Ω)/R.~~

~~For all vh ∈ Xh and qh ∈ Yh, require 1 h∇un+1, ∇v i − h∇un, ∇v i + νh∆un+1, ∆v i ∆t h h h h h h n n n n = h∇ph, ∆vhi − hf − uh · ∇uh, ∆vhi.~~

n n n n n h∇ph, ∇qhi = hf − uh ·∇uh, ∇qhi + νh∇ × uh, n×∇qhiΓ, Theorem Assume Ω is a bounded domain in RN (N=2,3) with C3 boundary. Let 0 M0, > 0, and let T∗ > 0 be given by the stability theorem. Let m ≥ 2, m ≥ 1 be integers, and assume

~~m+1 1 N m0 For any v ∈ H ∩ H0 (Ω, R ) and q ∈ H (Ω),~~

~~k−1 inf k∆(v − vh)k ≤ C0h kvkHk+1 for 2 ≤ k ≤ m, vh∈X0,h 32~~

~~m0−1 inf k∇(q − qh)k ≤ C0h kqkHm0, qh∈Yh~~

~~where C0 > 0 is independent of v, q and h.~~

~~0 Then there exists C1 > 0 with the following property. Whenever uh ∈ Xh, 0 < h < 1, 0 < n∆t ≤ min(T,T∗), and n 0 2 0 2 X 2 k∇uhk + ∆tk∆uhk + kf(tk)k ∆t ≤ M0, k=0~~

~~n n n n 1 then e = u(tn) − uh, r = p(tn) − ph of C ﬁnite element scheme satisfy~~

~~n X sup k∇ekk2 + k∆ekk2 + k∇rkk2 ∆t 0≤k≤n k=0 2 2m−2 2m0−2 0 2 0 2 ≤ C1(∆t + h + h + k∇e k + k∆e k ∆t). 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~~

~~un+1 − U n = ν∆un+1 + F n+1 in Ω, un+1 = 0 on ∂Ω, ∆t U n+1 = un+1 − ∇φn+1 = Pun+1~~

~~Why does it work? Apply P. Stokes pressure reappears implicitly:~~

~~U n+1 − U n + ν∇p (U n+1)= ν∆U n+1 + PF n+1, ∆t S U n+1 = −∇φn+1 on ∂Ω (slip error) 34~~

~~Subtract—Formally we get a consistent 1st-order scheme, since φn+1 − ν∆φn+1 = νp (U n+1) + Q(f n+1 − U n · ∇U n), ∆t S n · ∇φn+1 = 0 on ∂Ω.~~

~~n+1 thus ∇φ |∂Ω = O(∆t). The correct corresponding pressure~~

n+1 n+1 n+1 n+1 n+1 p = νpS(U ) + Q(f − U · ∇U ) φn+1 = − ν∆φn+1 + O(∆t) ∆t The (old) pressure approximation φn+1/∆t has a boundary layer with O(1) gradient error on ∂Ω, since n · ∇φn+1 = 0 but

~~n ·∇pn+1 = n · (ν∆U n+1 + f n+1 − U n+1 · ∇U n+1). 35 Higher-order accurate time discretization~~

~~Chorin-Temam method: 1st-order in time, boundary layers for p Brown, Cortez, Minion ’01: Explain 2nd-order time accuracy Guermond, Minev, Shen ’06: 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 ∇pn+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 A stable (3,2) slip-corrected projection method 1 X Use BD3 formula D un+1 = α un+1 + α un+1−k 3 ∆t 0 k k≥1

~~Write Hj = U j · ∇U j & extrapolate:~~

~~n+1 n n−1 n−2 n+1 n n−1 E3H = 3H − 3H + H , E2φ = 2φ − φ .~~

~~1. Given U j ≈ u(j∆t) (j ≤ n), solve (without any pressure!)~~

~~1 X α un+1 + α U n+1−k = ν∆un+1 + f n+1 − E Hn+1, ∆t 0 ∗ k ∗ 3 k≥1 n+1 n+1 n+1 u∗ = g + ∇E2φ on Γ.~~

~~n+1 n+1 n+1 n+1 n+1 n+1 2. U = u∗ − ∇φ , ∆φ = ∇ · u , n · ∇φ = 0. 37~~

~~Formal accuracy: Apply P and recall~~

~~j j j j Pu∗ = PU = U − ∇R(g ),~~

~~n+1 n+1 n+1 n+1 n+1 P∆u∗ = ∆Pu∗ − ∇pS(u∗ ) = ∆U −∇ pS(U ):~~

~~n+1 n+1 n+1 n+1 n+1 D3U + ∇p¯ = ν∆U + f − E3H ,~~

~~n+1 n+1 n+1 n+1 n+1 p¯ = νpS(U ) − R(D3g ) + Qf − QE3H ,~~

n+1 n+1 n+1 n+1 U = g − ∇(φ − E2φ ) on Γ. α 0 − ν∆ (φn+1 − E φn+1) =p ¯n+1 − 2¯pn +p ¯n−1 = O(∆t2), ∆t 2 hence U n+1 − gn+1 = O(∆t3) formally and the scheme is formally 3rd-order accurate overall. 38 A stable (3,2) pressure approximation method

~~n n n n Update intermediate velocity u∗ via (H = u∗ · ∇u∗ ) 1 X F = f n+1 − E Hn+1 − α un+1−j, 3 ∆t j ∗ j≥1 α P¯ = νp (E un+1) − 0 R(gn+1) + QF S 2 ∗ ∆t~~

~~α 0 un+1 − ν∆un+1 = −∇P¯ + F ∆t ∗ ∗ n+1 n+1 u∗ = g on Γ~~

The div-free velocity and consistent pressure are n+1 n+1 n+1 n+1 n+1 n+1 U = u∗ − ∇φ , ∆φ = ∇ · u , n · ∇φ = 0, n+1 ¯ n+1 p = P − ν∇ · u∗ . (Improves over previous methods) 39 (3,3) pressure update method (cf. BCG ’89, Ren et al ’05)

~~n+1 1. Find u∗ to solve~~

~~1 X α un+1 + α U n+1−k + ∇E pn+1 ∆t 0 ∗ k 3 k≥1 n+1 n+1 n+1 = ν∆u∗ + f − E3H ,~~

~~n+1 n+1 u∗ = g on Γ.~~

n+1 n+1 n+1 n+1 n+1 n+1 2. U = u∗ − ∇φ , ∆φ = ∇ · u , n · ∇φ = 0. 3. Update pressure via α pn+1 = E pn+1 + 0 − ν∆ φn+1 3 ∆t 40 Single mode Stokes test problem in strip −1 < x < 1

~~∂tu + ∇pS(u) = ∆u, u|x=±1 = 0~~

~~u(t, x, y) = eiξy−σt(u(x, ξ), iv(x, ξ))~~

For the case of odd symmetry: p = eiξy−σt sinh ξx, cosh ξx cos µx ξ cosh ξ u(x) = A − ,A = , cosh ξ cos µ ξ2 + µ2 sinh ξx sin µx ξ sinh ξ v(x) = B − ,B = . sinh ξ sin µ ξ2 + µ2

~~ξ tanh ξ + µ tan µ = 0, σ = ξ2 + µ2. Usually take ξ = 1, µ ≈ 2.883356, σ ≈ 9.314. 41 Single-mode time-discrete stability tests~~

~~n+1 n n+1−k A0u + A1u + ... + Aku = 0.~~

Look for un = κnu with         0 I 0 u I 0 0 u          0 0 I   κu  = κ  0 I 0   κu  . 2 2 −A3 −A2 −A1 κ u 0 0 A0 κ 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 Slip-correction schemes, single-mode stability test~~

~~1.1~~

~~0.9 (3,2) P1/P1 0.8 (3,2) P2/P1 (2,2) P1/P1 0.7 (2,2) P2/P1 (4,3) P1/P1 0.6 (4,3) P2/P1 0.5 −5 −3 −1 1 3 5 7 9 10 10 10 10 10 10 10 10 ∆t~~

1: max |κ| vs ∆t. 30 elements for each var. Solid lines: space-continuous theory. (2,2), (3,2): unconditional stability. (cf. Leriche et al ’06 w/PA) (3,3), (4,3): stability for ∆t < ∆tc independent of h, ξ. (smooth?) 43 Pressure update schemes, single-mode stability test

~~1.6 (3,2) P1/P1 (3,2) P2/P1 1.4 (2,2) P1/P1 (2,2) P2/P1 (3,3) P1/P1 1.2 (3,3) P2/P1 (4,3) P1/P1 (4,3) P2/P1 1~~

~~0.8~~

~~0.6~~

~~−5 −3 −1 1 3 5 7 9 10 10 10 10 10 10 10 10 ∆t~~

2: max |κ| vs ∆t. 30 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 2D test problem: stability & accuracy

~~Ω = square [−1, 1]2− hole, or ring. ν = 2, t = 2~~

~~ cos2(πx/2) sin(πy) u(t) = g(t) , − sin(πx) cos2(πy/2)~~

~~p(t) = g(t) cos(πx/2) sin(πy/2). 45 2D test: stability in domains with & without corners~~

∆tc \ N 0 1 2 (3,3) P1/P1 0.045 0.015 0.00067 (3,3) P2/P2 0.00027 0.000073 0.000018 (4,3) P1/P1 0.048 0.017 0.00074 (4,3) P2/P2 0.00029 0.000080 0.000020 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.

∆tc \ N 0 1 2 (3,3) P1/P1 0.022 0.014 0.010 (3,3) P2/P2 0.007 0.005 0.005 (4,3) P1/P1 0.024 0.017 0.014 (4,3) P2/P2 0.011 0.011 0.010 Largest time step for linear stability of PA schemes in annulus. N = # of refinements from grid in figure. No diffusive constraint on ∆t. 46 Accuracy: slip-corrected P4 FEM, square with hole

~~− log10 E (& local order α) vs. ∆t E \ ∆t 0.02 0.01 0.005~~

kp − pˆhkL2 4.18 (2.71) 5.08 (2.98) 5.99 (3.02) k∇(p − pˆh)kL2 3.47 (3.73) 4.39 (3.04) 5.17 (2.62) ku − uhkL∞ 5.75 (2.73) 6.65 (3.02) 7.57 (3.05) k∇(u − uh)kL∞ 3.94 (3.22) 4.95 (3.35) 5.83 (2.92) Top: (3,2) SC scheme. Bottom: (2,1) SC scheme.

~~E \ ∆t 0.02 0.01 0.005~~

kp − pˆhkL2 2.04 (1.97) 2.64 (2.02) 3.26 (2.04) k∇(p − pˆh)kL2 1.55 (1.96) 2.15 (2) 2.76 (2.02) ku − uhkL∞ 3.61 (1.96) 4.22 (2.02) 4.84 (2.05) k∇(u − uh)kL∞ 2.19 (1.97) 2.8 (2.01) 3.41 (2.04) 47 Pressure error for square with hole, (3,2) schemes

~~−5 −4 x 10 x 10 4 2~~

~~2 1~~

~~0 0~~

~~−2 −1~~

~~−4 −2 0.5 0.5 0.5 0.5 0 0 0 0 −0.5 −0.5 −0.5 −0.5~~

~~Pressure error for the (3,2) PA (left) and SC (right) schemes.~~

~~∆t = 0.02. 1296 P4 elements for each variable. Only values at vertices of triangles are used in the plots. 48~~

~~Benchmark tests with equal-order C0 ﬁnite elements~~

~~(u,v)=(1,0)~~

~~(u,v) given (u,v) given~~

~~3: Mesh used for backward facing step ﬂow when ν = 1/600 and for ﬂow past a cylinder when ν = 1/1000. 49 Driven Cavity, Re=1000, t=50~~

~~1 1 0.17 2 0 −5 −3−4 −2 0.12 3 −5 1 −2 3 −1 0.11~~

~~−4 0.11 −3 0.5 −0.5 0.07 0.09 1 −1 0.09 −2 −0.5 2 0.05 0.07 0.8 0.5 0.8 −5 3 0.05~~

~~−2 −3~~

~~0 1 −4 0.02 0 −0.002 0.6 0.6 0.02 1 0.07 2~~

~~−0.5~~

~~0.5 −2 3 0.05 −2 −1 −1 0 0.02 0.07~~

~~−2 0.4 −3 −0.5 0.4 0.5 1 2~~

~~3 0 0.09 1 0.05 0 0.11 0.09 0.2 1 1 −2 0.2 0.07 −0.5 −1 2 3 0.5 −0.5 0.5 0.11 0.5 2 2 0 0 0 3 1 0 0 0.2 0.4 0.6 0.8 1 0 0.2 0.4 0.6 0.8 1~~

2 × 64 × 64 stretched rectangular grid 8192 piecewise linear C0 elements for each variable. hmin = 0.00594, ∆t = 0.006, (3,2) slip-corrected scheme. Left: vorticity contours. Right: pressure contours. 50 Backward-facing step

~~0.5~~

~~−0.5 0 0.5 1 1.5 2 2.5 3 X~~

~~(3,2) SC scheme, 6640 P1 elements for each variable (dof=3487). ν = 1/100, t = 20, hmin = 0.00783, ∆t = 0.006.~~

~~X 1 X 0.5 2~~

~~0 S −0.5 X 0 1 2 3 4 5 3 6 7 8 9 10~~

~~(3,2) SC scheme, 1700 P2 elements for each variable (dof=3925). ν = 1/600, t = 120, hmin = 0.0186, ∆t = 0.003. 51 Flow past cylinder, ν = 1/1000~~

~~0.4~~

~~0.2~~

~~0 0 0.5 1 1.5 2~~

~~0.4~~

~~0.2~~

~~0 0 0.5 1 1.5 2~~

~~0.4~~

~~0.2~~

~~0 0 0.5 1 1.5 2~~

~~0.4~~

~~0.2~~

~~0 0 0.5 1 1.5 2~~

~~0.4~~

~~0.2~~

~~0 0 0.5 1 1.5 2~~

~~0.4~~

~~0.2~~

~~0 0 0.5 1 1.5 2 52 Drag, lift, pressure drop~~

~~t(c )=3.9348, c =2.9293 t(c )=5.6932, c =0.47756 ∆p =2.3219, ∆p(8)=−0.11148 d,max d,max l,max l,max max 3 0.5 2.5~~

~~0.4 2.5 2 0.3~~

~~2 0.2 1.5 0.1 1.5 0 1 1 −0.1 0.5 0.5 −0.2 −0.3 0 0 −0.4~~

~~−0.5 −0.5 −0.5 0 2 4 6 8 0 2 4 6 8 0 2 4 6 8 t t t~~

~~(3,2) SC scheme~~

~~763 isoparametric P4 elements (dof=6322) for each variable. 53 Summary~~

~~• The Navier-Stokes pressure for strong solutions in C3 domains is determined by the current velocity and data by~~

~~p = νpS(u) − R(∂tg) + Q(f − u · ∇u).~~

~~• The Laplace-Leray commutator is ∇pS(u) and is strictly controlled by ∆u at leading order, when u = 0 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 6= 0 with simple proof 54~~

~~Thank you!~~