Navier–Stokes equations with Navier boundary
conditions in nearly flat domains
Luan T Hoang∗ and George R Sell
∗ School
of Mathematics, University of Minnesota
www.math.umn.edu/∼lthoang/
lthoang@math.umn.edu
April 9, 2007
Dynamical Systems and PDE Seminars
School Mathematics, University of Minnesota
L. Hoang-G. Sell (University of Minnesota)
NSE with Navier boundary conditions
April 9, 2007
1 / 18
Outline
1
Introduction
2
Main results
3
Settings
4
Boundary conditions
5
Non-linear estimate
6
Global sotutions
L. Hoang-G. Sell (University of Minnesota)
NSE with Navier boundary conditions
April 9, 2007
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Introduction
Navier-Stokes equations (NSE) in R3 with a potential body force

∂u


 ∂t + (u · ∇)u − ν∆u = −∇p + f ,
div u = 0,



u(x, 0) = u0 (x),
ν > 0 is the kinematic viscosity,
u = (u1 , u2 , u3 ) is the unknown velocity field,
p ∈ R is the unknown pressure,
f (t) is the body force,
u0 is the initial velocity.
L. Hoang-G. Sell (University of Minnesota)
NSE with Navier boundary conditions
April 9, 2007
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Navier boundary conditions
On the boundary ∂Ω:
u · N = 0,
ν[D(u)N]tan = 0,
where N is the outward normal vector, [ · ]tan denotes the tangential part,
D(u) =
1
∇u + (∇u)∗ .
2
We assume: ν = 1.
Note: if the bounadry is flat, say, part of x3 = const, then the
condiontions become the free boundary condition
u3 = 0,
∂3 u1 = ∂3 u2 = 0.
Works by Temam-Ziane, primitive equations.
L. Hoang-G. Sell (University of Minnesota)
NSE with Navier boundary conditions
April 9, 2007
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Thin domains
Ω = Ωε = {(x1 , x2 , x3 ) : (x1 , x2 ) ∈ T2 , h0ε (x1 , x2 ) < x3 < h1ε (x1 , x2 )},
where ε ∈ (0, 1],
h0ε = εg0 ,
h1ε = εg1 ,
and g0 , g1 are given C 4 functions defined on T2 .
The boundary is Γ = Γ0 ∪ Γ1 , where Γ0 is the bottom and Γ1 is the top.
We define
1
M0 φ(x ) =
h1 − h0
0
Z
h1
φ(x 0 , x3 )dx3 ,
h0
b = (M0 u1 , M0 u2 , 0).
Mu
L. Hoang-G. Sell (University of Minnesota)
NSE with Navier boundary conditions
April 9, 2007
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Main results
Theorem (Global Existence Theorem)
There are κ > 0 and ε0 > 0 such that if 0 < ε ≤ ε0 , u0 ∈ V and f :
b 0 k2 2 ≤ κ 2 ,
ku0 k2H 1 ≤ κ2 ε−1 , kMu
L
2
2 −1
b
kPf kL∞ L2 ≤ κ ε , kM(Pf
)k2L∞ L2 ≤ κ2
(1)
then there exists a unique, globally defined strong solution u = u(t) of the
Navier-Stokes equations, with u(0) = u0 :
u ∈ C 0 ([0, ∞); H 1 (Ωε )) ∩ L∞ ((0, ∞); H 1 (Ωε )) ∩ L2loc ([0, ∞); H 2 (Ωε )).
Also, one has
ku(t)k2H 1 ≤ ε−1 M12 e −2αt + L21 ,
for all t ≥ 0,
where M1 , L1 , α > 0.
L. Hoang-G. Sell (University of Minnesota)
NSE with Navier boundary conditions
April 9, 2007
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Theorem (Global Attractor)
Suppose f is independent of t, and f satisfies the above condition. Then
the global attractor A for the above strong solutions also attracts all
Leray-Hopf weak solutions of the Navier–Stokes equations .
L. Hoang-G. Sell (University of Minnesota)
NSE with Navier boundary conditions
April 9, 2007
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A Green’s formula
Z
Z
∆u · v dx =
[−2(Du : Dv ) + (∇ · u)(∇ · v )] dx
Z
+
{2((Du)N) · v − (∇ · u)(v · N)} dσ.
Ω
Ω
∂Ω
If u is divergence-free and satisfies the Navier boundary condition, v is
tangential to the boundary then
Z
Z
− ∆u · v dx = 2 (Du : Dv ) dx.
Ω
L. Hoang-G. Sell (University of Minnesota)
Ω
NSE with Navier boundary conditions
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Uniform Korn inequality
Let H0 = {(a1 , a2 , 0) : a1 ∂1 g + a2 ∂2 g = 0}, where g = g1 − g0 .
Lemma
Let u ∈ H 1 (Ωε ) ∩ H0⊥ , u is tangential to the boundary of Ωε . Then
C1 kukH 1 ≤ kDukL2 ≤ C2 kukH 1 ,
for ε ∈ (0, 1], and C1 , C2 are positive constant independent of ε.
L. Hoang-G. Sell (University of Minnesota)
NSE with Navier boundary conditions
April 9, 2007
9 / 18
Boundary conditions
Lemma
Let τ be a tangential vestor field on the boundary. If u satisfies the Navier
boundary conditions then
∂N
∂u
·N +u·
= 0,
∂τ
∂τ
∂N
∂u
·τ =u·
.
∂N
∂τ
Combining with the trace theorem on the thin domain:
ku3 kL2 ≤ C εkukH 1 ,
k∂3 u1 kL2 ≤ C εkukH 2 ,
L. Hoang-G. Sell (University of Minnesota)
k∂3 u2 kL2 ≤ C εkukH 2 .
NSE with Navier boundary conditions
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The Stokes operator
Leray-Helmholtz decomposition
L2 (Ωε )3 = H ⊕ H0 ⊕ H1⊥ = H1 ⊕ H1⊥ ,
where H1⊥ = {∇φ : φ ∈ H 1 (Ωε )}.
D(A) = {u ∈ H 2 (Ωε ) ∩ H : u satisfies the Navier boundary conditions}.
Let P denotes the (Leray) projection on H. Then the Stokes operator is:
Au = −P∆u,
u ∈ D(A).
Lemma
If ε is small and u ∈ D(A) then
kAu + ∆ukL2 ≤ C1 (εk∇ukL2 + kukL2 ),
C2 kAukL2 ≤ kukH 2 ≤ C3 kAukL2 ,
where C1 , C2 , C3 are independent of ε.
L. Hoang-G. Sell (University of Minnesota)
NSE with Navier boundary conditions
April 9, 2007
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Also, if u ∈ V = D(A1/2 ) then
C1 kA1/2 ukL2 ≤ kukH 1 ≤ C5 kA1/2 ukL2 .
L. Hoang-G. Sell (University of Minnesota)
NSE with Navier boundary conditions
April 9, 2007
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Non-linear estimate
Proposition
For any ε ∈ (0, 1], u ∈ D(A) and β > 0, one has
|h(u · ∇)u, Aui| ≤ β kuk2H 2 + C ε1/2 kukH 1 kuk2H 2
+ Cβ ε−1 kuk2L2 kuk2H 1 ,
where C > 0 is independent of β and ε; and Cβ > 0 is independent of ε.
L. Hoang-G. Sell (University of Minnesota)
NSE with Navier boundary conditions
April 9, 2007
13 / 18
Non-linear estimate
Proposition
For any ε ∈ (0, 1], u ∈ D(A) and β > 0, one has
|h(u · ∇)u, Aui| ≤ β kuk2H 2 + C ε1/2 kukH 1 kuk2H 2
+ Cβ ε−1 kuk2L2 kuk2H 1 ,
where C > 0 is independent of β and ε; and Cβ > 0 is independent of ε.
Corollary
There is ε∗ ∈ (0, 1] such that for any ε < ε∗ , u ∈ D(A) and any β > 0 we
have
|h(u · ∇)u, Aui| ≤ β kAuk2L2 + C ε1/2 kA1/2 ukkAuk2L2
+ Cβ ε−1 kuk2L2 kA1/2 uk2 ,
where C > 0 is independent of β and ε; and Cβ > 0 is independent of ε.
L. Hoang-G. Sell (University of Minnesota)
NSE with Navier boundary conditions
April 9, 2007
13 / 18
Sketch of the proof
Averaging operator:
Mu = M0 u1 , M0 u2 , (M0 u1 , M0 u2 ) · ψ ,
where ψ is determined so that Mu ∈ H1 whenver u ∈ H1 .
Let v = Mu and w = u − v .
Establish Ladyzhenskaya inequality for v .
Establish Agmon, Poincare, Ladyzhenskaya, Galiardo-Nirengberg, Sobolev
inequalities for w .
hu · ∇u, Aui = hw · ∇u, Aui + hv · ∇u, Au + ∆ui − hv · ∇u, ∆ui.
Integration by parts: −hv · ∇u, ∆ui = I2 + I3 , where
Z
∂u
dσ.
I3 = (v · ∇)u
∂N
Γ
L. Hoang-G. Sell (University of Minnesota)
NSE with Navier boundary conditions
April 9, 2007
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Proof (continued)
Let b = (v · ∇)u = b1 τ1 + b2 τ2 + b3 N.
|b1 τ1 ·
∂N
∂u
| = |b1 u ·
| ≤ C ε|v ||∇u||u|.
∂N
∂τ
|b3 N ·
∂u
| ≤ C |b3 ||∇u|.
∂N
Noting that v = v1 τ1 + v2 τ2 ,
|b3 | = |(∇u)v · N| = | − (∇N)v · u)| = |u · (v1
Hence
∂N
∂N
+ v2
)| ≤ ε|u||v |.
∂τ1
∂τ2
Z
I3 ≤ ε
|u||v ||∇u|dσ.
Γ
L. Hoang-G. Sell (University of Minnesota)
NSE with Navier boundary conditions
April 9, 2007
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Global sotutions
b
Note k(I − M)uk
L2 ≤ C εkukH 1 .
L2 -Estimate.
b
b
b , Mui|
b
|hf , ui| = |h(I − M)Pf
, (I − M)ui
+ hMPf
b kL2 kukL2 .
= C εkPf kL2 kukH 1 + kMPf
Then
d
b k2 2 + C ε2 kPf k2 2 .
kuk2L2 + 2αkA1/2 uk2L2 ≤ C kMPf
L
L
dt
b 0 k2 2 + C ε2 ku0 k2 1 .
b 0 k2 2 + k(I − M)u
b 0 k2 2 ≤ kMu
ku0 k2L2 = kMu
L
H
L
L
ku(t)k2L2 ,
Z
L. Hoang-G. Sell (University of Minnesota)
t
t+1
kA1/2 uk2L2 ds ≤ C κ2 (e −2αt + 1).
NSE with Navier boundary conditions
April 9, 2007
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H 1 -estmimate
1d
kA1/2 uk2L2 +kAuk2L2 ≤ βkuk2H 2 +C ε1/2 kukH 1 kuk2H 2 +Cβ ε−1 kuk2L2 kuk2H 1 +k
2 dt
d
kA1/2 uk2L2 +(1−C ε1/2 kA1/2 ukL2 )kAuk2L2 ≤ C ε−1 kuk2L2 kA1/2 uk2L2 +C kPf k2L
dt
As far as 1 − C ε1/2 kA1/2 ukL2 < 1/2, say in [0, T ) then by the Uniform
Gronwall Inequality one has for 1 < t < T :
Z t
1/2
2
kA u(t)kL2 ≤
C ε−1 kuk2L2 kA1/2 uk2L2 +C kPf k2L2 +kA1/2 uk2L2 ds ≤ C κ2 ε−
t−1
L. Hoang-G. Sell (University of Minnesota)
NSE with Navier boundary conditions
April 9, 2007
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Acknowlegement. We thank Marta Lewicka for sharing her geometric
point of view on the boundary condition.
THANK YOU FOR YOUR ATTENTION.
L. Hoang-G. Sell (University of Minnesota)
NSE with Navier boundary conditions
April 9, 2007
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