1

Analysis and Computation of Navier-Stokes equation inbounded domains

Jian-Guo Liu

Department of Physics and Department of Mathematics

Duke University

with

Jie Liu (National University of Singapore)

Bob Pego (Carnegie Mellon)

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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

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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

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Leray-Helmholtz projection P onto divergence-free fields

L2(Ω,RN) = ∇H1(Ω)⊕ PL2(Ω,RN)

v = ∇φ+ w,

〈w,∇ψ〉 = 〈v −∇φ,∇ψ〉 = 0 for all ψ ∈ H1(Ω).

∆φ = ∇ · v, ∂nφ∣∣∣Γ

= n · v

∇ ·w = 0 in Ω, n ·w = 0 on ∂Ω.

Notation: w = Pv, φ = Qv.

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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

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

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pS arises from tangential vorticity at the boundary

Johnston-Liu

〈∇pS,∇q〉 = −〈n× (∇× u),∇q〉Γ, ∀q ∈ H1(Ω)

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Johnston-Liu

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Space of Stokes pressures

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

Sp = p ∈ H1(Ω)/R : ∆p = 0 in Ω, n ·∇p ∈ SΓ,

SΓ = f ∈ H−1/2(Γ) :

∫G

f = 0 ∀components G of Γ.

• ∃ a bounded right inverse ∇pS 7→ u from Sp→ H2 ∩H10(Ω,RN)

• In R3, ∇Sp is the space of simultaneous gradients and curls:

∇Sp = ∇H1(Ω) ∩ ∇×H1(Ω,R3)

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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∣∣∣Γ

= n · f − n · ∇ ×∇× u

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Pressure formula with inflow/outflow

Suppose we require: u = g on ∂Ω,

where to conserve volume,

∫∂Ω

n · g = 0 (t ≥ 0).

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:

〈∇p,∇q〉 = ν〈∇ × u,n×∇q〉Γ − 〈n·gt, q〉Γ + 〈f − u · ∇u,∇q〉

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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∣∣∣Γ

= n·(f−u·∇u−gt)−n·∇×∇×u

〈∇p,∇q〉 = ν〈∇ × u,n×∇q〉Γ − 〈n·gt, q〉Γ + 〈f − u · ∇u,∇q〉

〈∇p,∇q〉 = ν〈(∆−∇∇·)u,∇q〉 − 〈n·gt, q〉Γ + 〈f − u · ∇u,∇q〉〈n·gt, q〉Γ + 〈u · ∇u +∇p,∇q〉 = 〈ν∆u + f ,∇q〉 − ν〈∇∇ · u,∇q〉

〈(∇ · u)t, q〉+ ν〈∇(∇ · u),∇q〉 = 0

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

If (∇ · u)∣∣∣t=0

= 0, then ∇ · u ≡ 0. Equivalent to the original NSE

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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-Babuska-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

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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)

Assumeuin ∈ Huin := H1(Ω,RN),

f ∈ Hf := L2(0, T ;L2(Ω,RN)),

g ∈ Hg := H3/4(0, T ;L2(Γ,RN)) ∩ L2(0, T ;H3/2(Γ,RN))

∩g | ∂t(n · g) ∈ L2(0, T ;H−1/2(Γ)),g = uin (t = 0, x ∈ Γ), 〈n · ∂tg, 1〉Γ = 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).

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Estimate for the Laplace-Leray commutator

Recall [∆,P]u = (∆P − P∆)u = ∇pS

For period box [P,∆]u = 0

If ∇ · u = 0, then ‖[∆,P]u‖ = ‖((I − P)∆−∇∇·)u‖ ≤ ‖∆u‖.

Theorem Let Ω ⊂ RN (N ≥ 2) be a bounded C3 domain Then

∀ε > 0 ∃C ≥ 0 such that for all u ∈ H2 ∩H10(Ω,RN),∫

Ω

|(∆P − P∆)u|2 ≤(

1

2+ ε

)∫Ω

|∆u|2 + C

∫Ω

|∇u|2

Hence our unconstrained NSE is fully dissipative:

ut + P(u · ∇u− f) + ν∇pS = ν∆u

NSE as perturbed heat equation!

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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

∆u = −2k sin(kx)e−ky = 2∂xp,

∇ · (u, 0) = 0,∇× (0, v) = 0 at y = 0 (boundary properties )

‖px‖2 = ‖py‖2 = πk2 (equal partition)⟨

∆u− ∂xp0− ∂yp

,∂xp

∂yp

⟩= ‖px‖2 − ‖py‖2 = 0 (orthogonality)

This implies: ‖∇p‖2 =1

2‖∆u‖2 ≤ 1

2(‖∆u‖2 + ‖∆v‖2)

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Basic ingredients in the (elementary) proof:

Let Φ(x) = dist(x,Γ), n(x) = −∇Φ(x)

Ωs = x ∈ Ω | Φ(x) < s, 〈f, g〉s := 〈f, g〉ΩsChoose a cutoff function ξ so ξ(x) = 1 (Φ < 1

2s), ξ(x) = 0 (Φ > s).

Decompose velocity field into parts parallel and normal to Γ:

u = u⊥ + u‖, u⊥ = ξnntu+ (1− ξ)u, u‖ = ξ(I−nnt)u.

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

satisfying on ∂Ω: ∇ · u‖ = 0, ∇× u⊥ = 0

(ii) An orthogonality identity: ‖∆u‖‖2 = ‖∇pS‖2 + ‖∇pS −∆u‖‖2

(iii) Sharp N2D estimates on tubes: ‖∇pS‖‖2s ∼ ‖∇pS⊥‖2s

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∇pS = (∆P − P∆)u = (I − P)(∆−∇∇·)(u‖ +

HHHH

u⊥)

(due to ∇× u⊥ = 0 on ∂Ω)

∇pS = ∆u‖ −∇∇ · u‖ − P(∆−∇∇·)u‖

〈∇pS,∇pS〉 = 〈∇pS,∆u‖〉 − 〈∇pS,∇∇ · u‖〉 − 〈∇pS,Pv〉= 〈∇pS,∆u‖〉 − 0− 0

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

‖∆u‖‖2 = ‖∇pS‖2 + ‖∇pS −∆u‖‖2 (orthogonality)

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‖∇p−∆u‖‖2 = ‖∇p‖2Ωcs + ‖(∇p−∆u‖)⊥‖2s + ‖(∇p−∆u‖)‖‖2s≥ ‖∇p‖2Ωcs + (1− ε)‖∇p⊥‖2s + ‖(∇p−∆u‖)‖‖2s + · · ·

0 = 〈(∇p−∆u‖)‖,∇p‖〉s + 〈(∇p−∆u‖)⊥,∇p⊥〉s +

∫Φ>s

|∇p|2

Hence

2〈(∇p−∆u‖)⊥,∇p⊥〉s ≤ −2〈(∇p−∆u‖)‖,∇p‖〉s≤ ‖(∇p−∆u‖)‖‖2s + ‖∇p‖‖2s

‖(∇p−∆u‖)‖‖2s ≥ ‖∇p‖‖2s + 2(‖∇p⊥‖2s − ‖∇p‖‖2s)− junk

‖∇p−∆u‖‖2 ≥ (1− ε)‖∇p‖2 + 2(‖∇p⊥‖2s − ‖∇p‖‖2s)− junk

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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∣∣∣∣∫Φ<s

|n·∇p|2 − |(I − nnt)∇p|2∣∣∣∣ ≤ C0s

∫Φ<s0

|∇p|2

In the limit s→ 0, it reduce to∣∣∣∣∫Γ

|n·∇p|2 −∫

Γ

|(I − nnt)∇p|2∣∣∣∣ ≤ C0

∫Ω

|∇p|2

In a half space: ‖n·∇p‖2 = ‖(I − nnt)∇p‖2.

Known as Rellich identity in 2D circular disk.

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‖∆u‖2 = ‖∆u‖‖2 + 2〈∆u‖,∆u⊥〉+ ‖∆u⊥‖2

≥ (1− ε)‖∆u‖‖2 − C‖∇u‖2

= (1− ε)(‖∇pS‖2 + ‖∇pS −∆u‖‖2)− C‖∇u‖2

≥ (2− 3ε)‖∇pS‖2 − C‖∇u‖2

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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

exists u ∈ H2 ∩H10(Ω,RN) such that∫

Ω

|∇pS(u)|2 > β

∫Ω

|∆u|2 + C

∫Ω

|∇u|2

Open Q: Is extended Navier-Stokes dynamics well-posed in Lipschitz

(or polygonal) domains?

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Difference scheme with pressure explicit (Johnston-L, 2004)

∆pn = ∇ · (fn − un · ∇un) in Ω

n ·∇pn = n · fn − νn · (∇×∇× un) on ∂Ω

un+1 − un

∆t− ν∆un+1 = fn − un · ∇un −∇pn

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)

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Time-discrete scheme with pressure explicit(related to projection methods: Ti96,Pe01,GS03,JL04)

un+1 − un

∆t− ν∆un+1 = fn − un · ∇un −∇pn,

un+1 = 0 on ∂Ω,

∇pn = ν∇pnS

+ (I − P)(fn − un · ∇un).

We have the pressure estimate, for fixed constant β < 1

‖∇pn‖ ≤ ν‖∇pnS‖+ ‖fn − un · ∇un‖

≤ νβ‖∆un‖+ ‖fn − un · ∇un‖+ C‖∇un‖

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Stability analysis in H10 : dot with −∆un+1

un+1 − un

∆t− ν∆un+1 = fn − un · ∇un −∇pn

‖∇un+1 −∇un‖2 + ‖∇un+1‖2 − ‖∇un‖2

2∆t+ ν‖∆un+1‖2

≤ ‖∆un+1‖(2‖fn − un · ∇un‖+ ν‖∇pnS‖)

≤ (ε1

2+ν

2)‖∆un+1‖2 +

2

ε1‖fn − un · ∇un‖2 +

ν

2‖∇pS‖2

This gives ‖∇un+1‖2 − ‖∇un‖2

∆t+ (ν − ε1)‖∆un+1‖2

≤ 8

ε1(‖fn‖2 + ‖un · ∇un‖2) + ν‖∇pn

S‖2

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Handling the pressure

‖∇un+1‖2 − ‖∇un‖2

∆t+ (ν − ε1)‖∆un+1‖2

≤ 8

ε1(‖fn‖2 + ‖un · ∇un‖2) + ν‖∇pn

S‖2

Use the theorem (β = 12 + ε): ν‖∇pn

S‖2 ≤ νβ‖∆un‖2 + νC‖∇un‖2

‖∇un+1‖2 − ‖∇un‖2

∆t+ (ν − ε1)(‖∆un+1‖2 − ‖∆un‖2)

+ (ν − ε1 − νβ)‖∆un‖2

≤ 8

ε1(‖fn‖2 + ‖un · ∇un‖2) + νC‖∇un‖2

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Handling the nonlinear term

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

‖un · ∇un‖2 ≤ ε2‖∆un‖2 +4C

ε2‖∇un‖6

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

‖∇un+1‖2 − ‖∇un‖2 + ε‖∆un‖2

≤ C(‖fn‖2 + ‖∇un‖2 + ‖∇un‖6)

An easy discrete Gronwall inequality leads to:

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Stability theorem for N = 2, 3

Theorem Take f ∈ L2(0, T ;L2(Ω,RN)), u0 ∈ H2 ∩H10(Ω,RN).

Then ∃ positive constants T ∗ and C∗ depending only upon Ω, ν and

M0 := ‖∇u0‖2 + ν∆t‖∆u0‖2 +

∫ T

0

‖f‖2,

so that whenever n∆t ≤ T ∗ we have

sup0≤k≤n

‖∇uk‖2 +

n−1∑k=0

(‖∆uk‖2 +

∥∥∥∥uk+1 − uk

∆t

∥∥∥∥2)

∆t ≤ C∗.

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• Classic inf-sup condition for steady Stokes flow FEM

Find u ∈ X ⊂ H10(Ω,RN), p ∈ Y ⊂ L2(Ω)/R so that

〈∇u,∇v〉+ 〈∇p,v〉 = 〈f ,v〉 ∀v ∈ X,〈u,∇q〉 = 0 ∀q ∈ Y.

infp∈Y

supv∈X

〈p,∇ · v〉‖p‖‖∇v‖

≥ c > 0. (1)

Well-known:

• Condition (1) is necessary and sufficient for stability of

(p,u) in L2 ×H10 (the weak NS solution space).

• Many simple finite elements fail this condition

(e.g., equal-order Lagrange)

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Steady Stokes flow in terms of Stokes pressure

Find u ∈ X ⊂ H2 ∩H10(Ω,RN), p ∈ Y ⊂ H1(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) ∈ H1× (H2 ∩H10(Ω,RN)) (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

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

C0 finite element scheme: Xh ⊂ H10(Ω,RN), Yh ⊂ H1(Ω)/R. For

all vh ∈ Xh and qh ∈ Yh, we require require

〈∇pnh,∇qh〉 = 〈fn − unh ·∇unh,∇qh〉+ ν〈∇ × unh,n×∇qh〉Γ,

〈un+1h − unh

∆t,vh〉+ ν〈∇un+1

h ,∇vh〉 = 〈fn − unh · ∇unh −∇pnh,vh〉.

• 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).

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Error estimates of C1 FE scheme

unh ∈ Xh ⊂ H2 ∩H10(Ω,RN), pnh ∈ Yh ⊂ H1(Ω)/R.

For all vh ∈ Xh and qh ∈ Yh, require

1

∆t

(〈∇un+1

h ,∇vh〉 − 〈∇unh,∇vh〉)

+ ν〈∆un+1h ,∆vh〉

= 〈∇pnh,∆vh〉 − 〈fn − unh · ∇unh,∆vh〉.

〈∇pnh,∇qh〉 = 〈fn − unh ·∇unh,∇qh〉+ ν〈∇ × unh,n×∇qh〉Γ,

Theorem Assume Ω is a bounded domain in RN (N=2,3) with C3 boundary. LetM0, > 0, and let T∗ > 0 be given by the stability theorem. Let m ≥ 2, m′ ≥ 1 beintegers, and assume

For any v ∈ Hm+1 ∩H10(Ω,RN) and q ∈ Hm′(Ω),

infvh∈X0,h

‖∆(v − vh)‖ ≤ C0hk−1‖v‖Hk+1 for 2 ≤ k ≤ m,

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infqh∈Yh

‖∇(q − qh)‖ ≤ C0hm′−1‖q‖Hm′,

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

Then there exists C1 > 0 with the following property. Whenever u0h ∈ Xh,

0 < h < 1, 0 < n∆t ≤ min(T, T∗), and

‖∇u0h‖2 + ∆t‖∆u0

h‖2 +

n∑k=0

‖f(tk)‖2∆t ≤M0,

then en = u(tn)− unh, rn = p(tn)− pnh of C1 finite element scheme satisfy

sup0≤k≤n

‖∇ek‖2 +

n∑k=0

(‖∆ek‖2 + ‖∇rk‖2

)∆t

≤ C1(∆t2 + h2m−2 + h2m′−2 + ‖∇e0‖2 + ‖∆e0‖2∆t).

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3. Split-step time discretizations of NSE

Chorin/Temam projection method

Given velocity Un = PUn, let F n+1 = fn+1 −Un · ∇Un

and update velocity by solving

un+1 −Un

∆t= ν∆un+1 + F n+1 in Ω, un+1 = 0 on ∂Ω,

Un+1 = un+1 −∇φn+1 = Pun+1

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

Un+1 −Un

∆t+ ν∇pS(U

n+1) = ν∆Un+1 + PF n+1,

Un+1 = −∇φn+1 on ∂Ω (slip error)

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Subtract—Formally we get a consistent 1st-order scheme, since

φn+1

∆t− ν∆φn+1 = νpS(U

n+1) +Q(fn+1 −Un · ∇Un),

n · ∇φn+1 = 0 on ∂Ω.

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

pn+1 = νpS(Un+1) +Q(fn+1 −Un+1 · ∇Un+1)

=φn+1

∆t− ν∆φn+1 +O(∆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∆Un+1 + fn+1 −Un+1 · ∇Un+1).

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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 Un+1

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A stable (3,2) slip-corrected projection method

Use BD3 formula D3un+1 =

1

∆t

(α0u

n+1 +∑k≥1

αkun+1−k

)Write Hj = U j · ∇U j & extrapolate:

E3Hn+1 = 3Hn − 3Hn−1 + Hn−2, E2φn+1 = 2φn − φn−1.

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

1

∆t

(α0u

n+1∗ +

∑k≥1

αkUn+1−k

)= ν∆un+1

∗ + fn+1 − E3Hn+1,

un+1∗ = gn+1 +∇E2φn+1 on Γ.

2. Un+1 = un+1∗ −∇φn+1, ∆φn+1 = ∇ · un+1, n · ∇φn+1 = 0.

37

Formal accuracy: Apply P and recall

Puj∗ = PU j = U j −∇R(gj),

P∆un+1∗ = ∆Pun+1

∗ −∇pS(un+1∗ ) = ∆Un+1 −∇pS(U

n+1) :

D3Un+1 +∇pn+1 = ν∆Un+1 + fn+1 − E3Hn+1,

pn+1 = νpS(Un+1)−R(D3g

n+1) +Qfn+1 −QE3Hn+1,

Un+1 = gn+1 −∇(φn+1 − E2φn+1) on Γ.(α0

∆t− ν∆

)(φn+1 − E2φn+1) = pn+1 − 2pn + pn−1 = O(∆t2),

hence Un+1 − gn+1 = O(∆t3) formally and the scheme is formally

3rd-order accurate overall.

38

A stable (3,2) pressure approximation method

Update intermediate velocity un∗ via (Hn = un∗ · ∇un∗)

F = fn+1 − E3Hn+1 − 1

∆t

∑j≥1

αjun+1−j∗ ,

P = νpS(E2un+1∗ )− α0

∆tR(gn+1) +QF

α0

∆tun+1∗ − ν∆un+1

∗ = −∇P + F

un+1∗ = gn+1 on Γ

The div-free velocity and consistent pressure are

Un+1 = un+1∗ −∇φn+1, ∆φn+1 = ∇ · un+1, n · ∇φn+1 = 0,

pn+1 = P − ν∇ · un+1∗ . (Improves over previous methods)

39

(3,3) pressure update method (cf. BCG ’89, Ren et al ’05)

1. Find un+1∗ to solve

1

∆t

(α0u

n+1∗ +

∑k≥1

αkUn+1−k

)+∇E3pn+1

= ν∆un+1∗ + fn+1 − E3Hn+1,

un+1∗ = gn+1 on Γ.

2. Un+1 = un+1∗ −∇φn+1, ∆φn+1 = ∇ · un+1, n · ∇φn+1 = 0.

3. Update pressure via

pn+1 = E3pn+1 +(α0

∆t− ν∆

)φn+1

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,

u(x) = A

(cosh ξx

cosh ξ− cosµx

cosµ

), A =

ξ cosh ξ

ξ2 + µ2,

v(x) = B

(sinh ξx

sinh ξ− sinµx

sinµ

), B =

ξ sinh ξ

ξ2 + µ2.

ξ tanh ξ + µ tanµ = 0, σ = ξ2 + µ2.

Usually take ξ = 1, µ ≈ 2.883356, σ ≈ 9.314.

41

Single-mode time-discrete stability tests

A0un+1 +A1u

n + . . .+Akun+1−k = 0.

Look for un = κnu with 0 I 0

0 0 I

−A3 −A2 −A1

u

κu

κ2u

= κ

I 0 0

0 I 0

0 0 A0

u

κu

κ2u

.

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

10−5

10−3

10−1

101

103

105

107

109

0.5

0.6

0.7

0.8

0.9

1

1.1

∆t

(3,2) P1/P1(3,2) P2/P1(2,2) P1/P1(2,2) P2/P1(4,3) P1/P1(4,3) P2/P1

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

10−5

10−3

10−1

101

103

105

107

109

0.6

0.8

1

1.2

1.4

1.6

∆t

(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/P1

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

u(t) = g(t)

(cos2(πx/2) sin(πy)

− 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

− log10E (& local order α) vs. ∆t

E \ ∆t 0.02 0.01 0.005

‖p− ph‖L2 4.18 (2.71) 5.08 (2.98) 5.99 (3.02)

‖∇(p− ph)‖L2 3.47 (3.73) 4.39 (3.04) 5.17 (2.62)

‖u− uh‖L∞ 5.75 (2.73) 6.65 (3.02) 7.57 (3.05)

‖∇(u− uh)‖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 0.02 0.01 0.005

‖p− ph‖L2 2.04 (1.97) 2.64 (2.02) 3.26 (2.04)

‖∇(p− ph)‖L2 1.55 (1.96) 2.15 (2) 2.76 (2.02)

‖u− uh‖L∞ 3.61 (1.96) 4.22 (2.02) 4.84 (2.05)

‖∇(u− uh)‖L∞ 2.19 (1.97) 2.8 (2.01) 3.41 (2.04)

47

Pressure error for square with hole, (3,2) schemes

−0.5

0

0.5

−0.5

0

0.5−4

−2

0

2

4

x 10−5

−0.5

0

0.5

−0.5

0

0.5−2

−1

0

1

2

x 10−4

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 acylinder when ν = 1/1000.

49

Driven Cavity, Re=1000, t=50

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1−5 −5

−5

−4

−4

−4

−3

−3

−3

−3

−2

−2

−2

−2

−2

−2

−2−

2

−1

−1

−1

−1−

1

−0.5

−0.5

−0.5

−0.5

−0.

5−

0.5

0

0

0

0

00 0

0.5

0.5

0.5

0.5

0.5

0.5

0.5

1

1

1

1

1

1

1

1

1

2

2

2

2

2

2

2

23

3

3

33

33

0 0.2 0.4 0.6 0.8 10

0.2

0.4

0.6

0.8

1

−0.002

00.02

0.020.02

0.05

0.05

0.05

0.05

0.07

0.07

0.07

0.07

0.07

0.09

0.09

0.09 0.09

0.11

0.11

0.11

0.110.120.17

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 0.5 1 1.5 2 2.5 3−0.5

0

0.5

S

X

(3,2) SC scheme, 6640 P1 elements for each variable (dof=3487).

ν = 1/100, t = 20, hmin = 0.00783, ∆t = 0.006.

0 1 2 3 4 5 6 7 8 9 10−0.5

0

0.5

X3

X1 X

2

S

(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 0.5 1 1.5 20

0.2

0.4

0 0.5 1 1.5 20

0.2

0.4

0 0.5 1 1.5 20

0.2

0.4

0 0.5 1 1.5 20

0.2

0.4

0 0.5 1 1.5 20

0.2

0.4

0 0.5 1 1.5 20

0.2

0.4

52

Drag, lift, pressure drop

0 2 4 6 8−0.5

0

0.5

1

1.5

2

2.5

3

t(cd,max

)=3.9348, cd,max

=2.9293

t0 2 4 6 8

−0.5

−0.4

−0.3

−0.2

−0.1

0

0.1

0.2

0.3

0.4

0.5

t(cl,max

)=5.6932, cl,max

=0.47756

t0 2 4 6 8

−0.5

0

0.5

1

1.5

2

2.5

∆pmax

=2.3219, ∆p(8)=−0.11148

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!