Navier-Stokes equations in thin domains with Navier
friction boundary conditions
Luan Thach Hoang
Department of Mathematics and Statistics, Texas Tech University
www.math.umn.edu/∼lhoang/
luan.hoang@ttu.edu
Applied Mathematics Seminar
Texas Tech University
September 17, 24, 2008
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Navier friction boundary conditions
Applied Math Seminar, Sept. 17,24, 2008
Outline
1
Introduction
2
Main results
3
Functional settings
4
Linear and non-linear estimates
5
Global solutions
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Navier friction boundary conditions
Applied Math Seminar, Sept. 17,24, 2008
Introduction
Navier-Stokes equations for fluid dynamics:


∂t u + (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 given initial data.
L. Hoang- Texas Tech
Navier friction boundary conditions
Applied Math Seminar, Sept. 17,24, 2008
Navier friction boundary conditions
On the boundary ∂Ω:
u · N = 0,
ν[D(u)N]tan + γu = 0,
N is the unit outward normal vector
γ ≥ 0 denotes the friction coefficients
[ · ]tan denotes the tangential part
D(u) =
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1
∇u + (∇u)∗ .
2
Navier friction boundary conditions
Applied Math Seminar, Sept. 17,24, 2008
Remarks
ν = 0, γ = 0: Boundary condition for inviscid fluids
γ = ∞: Dirichlet condition.
γ = 0: Navier boundary conditions (without friction)
[Iftimie-Raugel-Sell](with flat bottom), [H.-Sell].
If the boundary is flat, say, part of x3 = const, then the conditions
become the Robin conditions (see [Hu])
u3 = 0,
u1 + γ∂3 u1 = u2 + γ∂3 u2 = 0.
Compressible fluids on half planes with Navier friction boundary
conditions [Hoff].
Assume ν = 1.
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Navier friction boundary conditions
Applied Math Seminar, Sept. 17,24, 2008
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 3 functions defined on T2 ,
g = g1 − g0 ≥ c0 > 0.
The boundary is Γ = Γ0 ∪ Γ1 , where Γ0 is the bottom and Γ1 is the top.
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Navier friction boundary conditions
Applied Math Seminar, Sept. 17,24, 2008
Boundary conditions on thin domains
The velocity u satisfies the Navier friction boundary conditions on Γ1 and
Γ0 with friction coefficients γ1 = γ1ε and γ0 = γ0ε , respectively.
Assumption
There is δ ∈ [0, 1], such that for i=0,1,
0 < lim inf
ε→0
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γiε
γiε
≤
lim
sup
< ∞.
δ
εδ
ε→0 ε
Navier friction boundary conditions
Applied Math Seminar, Sept. 17,24, 2008
Notation
Leray-Helmholtz decomposition
L2 (Ωε )3 = H ⊕ H ⊥
where
H = {u ∈ L2 (Ωε )3 : ∇ · u = 0 in Ωε , u · N = 0 on Γ},
H ⊥ = {∇φ : φ ∈ H 1 (Ωε )}.
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Navier friction boundary conditions
Applied Math Seminar, Sept. 17,24, 2008
Notation
Leray-Helmholtz decomposition
L2 (Ωε )3 = H ⊕ H ⊥
where
H = {u ∈ L2 (Ωε )3 : ∇ · u = 0 in Ωε , u · N = 0 on Γ},
H ⊥ = {∇φ : φ ∈ H 1 (Ωε )}.
Let V be the closure in H 1 (Ωε , R3 ) of u ∈ C ∞ (Ωε , R3 ) ∩ H that satisfies
the friction boundary conditions.
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Navier friction boundary conditions
Applied Math Seminar, Sept. 17,24, 2008
Notation
Leray-Helmholtz decomposition
L2 (Ωε )3 = H ⊕ H ⊥
where
H = {u ∈ L2 (Ωε )3 : ∇ · u = 0 in Ωε , u · N = 0 on Γ},
H ⊥ = {∇φ : φ ∈ H 1 (Ωε )}.
Let V be the closure in H 1 (Ωε , R3 ) of u ∈ C ∞ (Ωε , R3 ) ∩ H that satisfies
the friction boundary conditions.
Averaging operator:
1
M0 φ(x ) =
εg
0
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Z
h1
φ(x 0 , x3 )dx3 ,
b = (M0 u1 , M0 u2 , 0).
Mu
h0
Navier friction boundary conditions
Applied Math Seminar, Sept. 17,24, 2008
Main result
Theorem (Global strong solutions)
Let δ ∈ [2/3, 1]. There are ε0 > 0 and κ > 0 such that if ε ∈ (0, ε0 ] and
u0 ∈ V and f ∈ L∞ (L2 ) satisfy
b 0 k2 2 ,
mu,0 = kMu
mu,1 = εku0 k2H 1 ,
L
b k2 ∞ 2 , mf ,1 = εkf k2 ∞ 2 ,
mf ,0 = kMf
L L
L L
are smaller than κ, then the regular solution exists for all t ≥ 0:
u ∈ C ([0, ∞), H 1 (Ωε )) ∩ L2loc ([0, ∞), H 2 (Ωε )).
Remark: The condition on u0 is acceptable.
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Navier friction boundary conditions
Applied Math Seminar, Sept. 17,24, 2008
A Green’s formula
[Solonnikov-Šcǎdilov]
Z
Z
∆u · v dx =
[−2(Du : Dv ) + (∇ · u)(∇ · v )] dx
Ω
Ω Z
+
{2((Du)N) · v − (∇ · u)(v · N)} dσ.
∂Ω
If u is divergence-free and satisfies the Navier friction boundary conditions,
v is tangential to the boundary then
Z
Z
−
∆u · v dx = 2
Ωε
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Z
u · v dσ + 2γ1
(Du : Dv ) dx + 2γ0
Ωε
Z
Γ0
Navier friction boundary conditions
u · v dσ.
Γ1
Applied Math Seminar, Sept. 17,24, 2008
A Green’s formula
[Solonnikov-Šcǎdilov]
Z
Z
∆u · v dx =
[−2(Du : Dv ) + (∇ · u)(∇ · v )] dx
Ω
Ω Z
+
{2((Du)N) · v − (∇ · u)(v · N)} dσ.
∂Ω
If u is divergence-free and satisfies the Navier friction boundary conditions,
v is tangential to the boundary then
Z
Z
−
∆u · v dx = 2
Ωε
Z
u · v dσ + 2γ1
(Du : Dv ) dx + 2γ0
Ωε
Z
Γ0
u · v dσ.
Γ1
The right hand side is denoted by E (u, v ).
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Navier friction boundary conditions
Applied Math Seminar, Sept. 17,24, 2008
Uniform Korn inequality
Is E (·, ·) bounded and coercive in H 1 (Ωε )?
We need Korn’s inequality: kuk2H 1 (Ωε ) ≤ Cε E (u, u).
Lemma
There is ε0 > 0 such that for ε ∈ (0, ε0 ], u ∈ H 1 (Ωε ) ∩ H0⊥ and u is
tangential to the boundary of Ωε , one has
kuk2L2 ≤ C ε1−δ E (u, u),
C kuk2H 1 ≤ E (u, u) ≤ C 0 (k∇uk2L2 + ε1−δ kuk2L2 ),
where C , C 0 are positive constants independent of ε.
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Navier friction boundary conditions
Applied Math Seminar, Sept. 17,24, 2008
Stokes operator
Let P denotes the (Leray) projection on H. Then the Stokes operator is:
Au = −P∆u,
u ∈ DA ,
DA = {u ∈ H 2 (Ωε )3 ∩V : u satisfies the Navier friction boundary conditions}
For u ∈ DA , v ∈ V , one has
hAu, v i = E (u, v ).
Navier-Stokes equations:
du
+ Au + B(u, u) = Pf ,
dt
where B(u, v ) = P(u · ∇u).
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Navier friction boundary conditions
Applied Math Seminar, Sept. 17,24, 2008
Inequalities
For ε ∈ (0, ε0 ], one has the following:
If u ∈ V = D
1
A2
then
1
kukL2 ≤ C ε(1−δ)/2 kA 2 ukL2 ,
1
kukH 1 ≤ C kA 2 ukL2 ,
1
kA 2 ukL2 ≤ C (k∇ukL2 + ε(δ−1)/2 kukL2 ).
If u ∈ DA then
1
kA 2 ukL2 ≤ C ε(1−δ)/2 kAukL2 ,
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kukL2 ≤ C ε1−δ kAukL2 .
Navier friction boundary conditions
Applied Math Seminar, Sept. 17,24, 2008
Interpreting the boundary conditions
Lemma
Let τ be a tangential vector field on the boundary. If u satisfies the Navier
friction boundary conditions then one has on Γ that
∂u
∂N
· N = −u ·
,
∂τ
∂τ
∂N
∂u
·τ =u·
− 2γτ .
∂N
∂τ
One also has [Chueshov-Raugel-Rekalo]
N × (∇ × u) = 2N × {N × ((∇N)∗ u) − γN × u}.
Our case: |∇N| ∼ ε and γ ∼ εδ .
L. Hoang- Texas Tech
Navier friction boundary conditions
Applied Math Seminar, Sept. 17,24, 2008
Linear and Non-linear Estimates
Proposition
If ε ∈ (0, ε0 ] and u ∈ DA , then
kAu + ∆ukL2 ≤ C1 εδ k∇ukL2 + C1 εδ−1 kukL2 ,
C2 kAukL2 ≤ kukH 2 ≤ C3 kAukL2 .
L. Hoang- Texas Tech
Navier friction boundary conditions
Applied Math Seminar, Sept. 17,24, 2008
Linear and Non-linear Estimates
Proposition
If ε ∈ (0, ε0 ] and u ∈ DA , then
kAu + ∆ukL2 ≤ C1 εδ k∇ukL2 + C1 εδ−1 kukL2 ,
C2 kAukL2 ≤ kukH 2 ≤ C3 kAukL2 .
Proposition
There is ε0 > 0 such that for ε ∈ (0, ε0 ], α > 0 and u ∈ DA , one has
1
1
|hu · ∇u, Aui| ≤ {α + C ε1/2 kA 2 ukL2 }kAuk2L2 + Cα ε2δ kuk2L2 kA 2 uk4L2
2/3
1
1
+ Cα ε−1 εδ−2/3 kukL2 kA 2 uk2L2 + Cα ε−1 kuk2L2 kA 2 uk2L2 .
where the positive number Cα depends on α but not on ε.
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Navier friction boundary conditions
Applied Math Seminar, Sept. 17,24, 2008
Corollary
Suppose δ ∈ [2/3, 1], then there exists ε∗ ∈ (0, 1] such that for any ε < ε∗
and u ∈ DA , one has
n1
o
1
|hu · ∇u, Aui| ≤
+ d1 ε1/2 kA 2 ukL2 kAuk2L2
4
n
o 1
n
o
1
1
+ d2 kuk2L2 kA 2 uk2L2 kA 2 uk2L2 + d3 1 + kuk2L2 ε−1 kA 2 uk2L2 .
where positive constants d1 , d2 and d3 are independent of ε.
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Navier friction boundary conditions
Applied Math Seminar, Sept. 17,24, 2008
Key identity
Lemma
Let u ∈ DA and Φ ∈ H 1 (Ωε )3 . One has
Z
Z
(∇ × (∇ × u)) · Φdx =
Ωε
Ωε
(∇ × Φ) · (∇ × u + G (u))dx
Z
−
Φ · (∇ × G (u))dx.
Ωε
where
|G (u)| ≤ C εδ |u|,
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|∇G (u)| ≤ C εδ |∇u| + C εδ−1 |u|.
Navier friction boundary conditions
Applied Math Seminar, Sept. 17,24, 2008
Key identity
Lemma
Let u ∈ DA and Φ ∈ H 1 (Ωε )3 . One has
Z
Z
(∇ × (∇ × u)) · Φdx =
Ωε
Ωε
(∇ × Φ) · (∇ × u + G (u))dx
Z
−
Φ · (∇ × G (u))dx.
Ωε
where
|G (u)| ≤ C εδ |u|,
|∇G (u)| ≤ C εδ |∇u| + C εδ−1 |u|.
Linear estimate: Φ = Au + ∆u, ∇ × Φ = 0.
Non-linear estimate: Φ = u × (∇ × u).
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Navier friction boundary conditions
Applied Math Seminar, Sept. 17,24, 2008
Estimate of k∇2 ukL2
Lemma
There is ε0 ∈ (0, 1] such that if ε < ε0 and u ∈ H 2 (Ωε )3 satisfies the
Navier friction boundary conditions, then
k∇2 ukL2 ≤ C k∆ukL2 + C kukH 1 .
Remarks on the proof. Integration by parts
Z
Z
Z 1 ∂|∇u|2
∂u
|∇2 u|2 dx =
|∆u|2 dx +
−
· ∆u dσ.
∂N
Ωε
Ωε
Γ 2 ∂N
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Navier friction boundary conditions
Applied Math Seminar, Sept. 17,24, 2008
Estimate of k∇2 ukL2
Lemma
There is ε0 ∈ (0, 1] such that if ε < ε0 and u ∈ H 2 (Ωε )3 satisfies the
Navier friction boundary conditions, then
k∇2 ukL2 ≤ C k∆ukL2 + C kukH 1 .
Remarks on the proof. Integration by parts
Z
Z
Z 1 ∂|∇u|2
∂u
|∇2 u|2 dx =
|∆u|2 dx +
−
· ∆u dσ.
∂N
Ωε
Ωε
Γ 2 ∂N
Remove the second derivatives in the boundary integrals
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Navier friction boundary conditions
Applied Math Seminar, Sept. 17,24, 2008
Estimate of k∇2 ukL2
Lemma
There is ε0 ∈ (0, 1] such that if ε < ε0 and u ∈ H 2 (Ωε )3 satisfies the
Navier friction boundary conditions, then
k∇2 ukL2 ≤ C k∆ukL2 + C kukH 1 .
Remarks on the proof. Integration by parts
Z
Z
Z 1 ∂|∇u|2
∂u
|∇2 u|2 dx =
|∆u|2 dx +
−
· ∆u dσ.
∂N
Ωε
Ωε
Γ 2 ∂N
Remove the second derivatives in the boundary integrals
Appropriate order for ε
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Navier friction boundary conditions
Applied Math Seminar, Sept. 17,24, 2008
Estimate of k∇2 ukL2
Lemma
There is ε0 ∈ (0, 1] such that if ε < ε0 and u ∈ H 2 (Ωε )3 satisfies the
Navier friction boundary conditions, then
k∇2 ukL2 ≤ C k∆ukL2 + C kukH 1 .
Remarks on the proof. Integration by parts
Z
Z
Z 1 ∂|∇u|2
∂u
|∇2 u|2 dx =
|∆u|2 dx +
−
· ∆u dσ.
∂N
Ωε
Ωε
Γ 2 ∂N
Remove the second derivatives in the boundary integrals
Appropriate order for ε
The role of the positivity of the friction coefficients
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Navier friction boundary conditions
Applied Math Seminar, Sept. 17,24, 2008
Estimate of k∇2 ukL2
Lemma
There is ε0 ∈ (0, 1] such that if ε < ε0 and u ∈ H 2 (Ωε )3 satisfies the
Navier friction boundary conditions, then
k∇2 ukL2 ≤ C k∆ukL2 + C kukH 1 .
Remarks on the proof. Integration by parts
Z
Z
Z 1 ∂|∇u|2
∂u
|∇2 u|2 dx =
|∆u|2 dx +
−
· ∆u dσ.
∂N
Ωε
Ωε
Γ 2 ∂N
Remove the second derivatives in the boundary integrals
Appropriate order for ε
The role of the positivity of the friction coefficients
k∇2 uk2L2 ≤ k∆uk2L2 + C kuk2H 1 + C ε2 k∇2 uk2L2 .
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Navier friction boundary conditions
Applied Math Seminar, Sept. 17,24, 2008
Estimate of the non-linear term
Write u = v + w where
1
{(x3 − h0 )∇2 h1 + (h1 − x3 )∇2 h0 } .
εg
Then v is divergence free and tangential to the boundary.
Important properties:
v is a 2D-like vector field.
w satisfies “good” inequalites:
b Mu
b · ψ),
v = Mu = (Mu,
ψ(x) =
1/2
1/2
kv kL4 ≤ C ε−1/4 kukL2 kukH 1 ,
kw kL2 ≤ C εk∇w kL2 ,
1/2
1/2
k∇v kL4 ≤ C ε−1/4 kukH 1 kukH 2
k∇w kL2 ≤ C εkukH 2 + C εδ kukL2 ,
1/2
1/2
kw kL∞ ≤ C ε1/2 kukH 2 + C εδ/2 kukL2 kukH 2 .
Then write
h(u · ∇)u, Aui = h(w · ∇)u, Aui + h(v · ∇)u, Au + ∆ui − h(v · ∇)u, ∆ui.
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Navier friction boundary conditions
Applied Math Seminar, Sept. 17,24, 2008
Strong global solutions
Do not need u = (v , w ) and equations for each v and w
Non-linear estimate and Uniform Gronwall’s inequality
Steps:
Estimates for ku(t)kL2 and
Rt
t−1 kA
1
2
u(s)k2L2 ds
1
Estimates for kA 2 u(t)k2L2 and the “right” ε−1 size.
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Navier friction boundary conditions
Applied Math Seminar, Sept. 17,24, 2008
L2 -Estimates for u
b
Poincaré-like inequalities: k(I − M)uk
L2 ≤ C εkukH 1 .
1
1d
kuk2L2 + kA 2 uk2L2 ≤ |hu, Pf i| ≤ |hM̂u, M̂Pf i| + |h(I − M̂)u, (I − M̂)Pf i|
2 dt
1 1
b k2 ∞ 2 + ε2 kPf k2 ∞ 2 .
≤ kA 2 uk2L2 + kMPf
L L
L L
2
Hence Gronwall’s inequality yields
b k2 ∞ 2 + ε2 kPf k2 ∞ 2 .
ku(t)k2L2 ≤ ku0 k2L2 e −c1 t + kMPf
L L
L L
b 0 k2 2 ≤ kMu
b 0 k2 2 + C ε2 ku0 k2 1 ≤ C κ.
b 0 k2 2 + k(I − M)u
ku0 k2L2 = kMu
H
L
L
L
ku(t)k2L2 ,
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Z
t
t+1
1
kA 2 uk2L2 ds ≤ C κ.
Navier friction boundary conditions
Applied Math Seminar, Sept. 17,24, 2008
H 1 -Estimate for u
o
n1
1
1
1d
kA 2 uk2L2 + kAuk2L2 ≤
+ d1 ε1/2 kA 2 ukL2 kAuk2L2
2n
dt
4 n
o 1
o
1
1
2
2
2
+d2 kukL2 kA 2 ukL2 kA 2 ukL2 +d3 1+kuk2L2 ε−1 kA 2 uk2L2 +kAukL2 kf kL∞ L2 .
1
1
1
d
kA 2 uk2L2 + (1 − 2d1 ε 2 kA 2 ukL2 )kAuk2L2
dt
1
≤ g kA 2 uk2L2 + h,
where
1
g = 2d2 kuk2L2 kA 2 uk2L2 ,
n
o
1
h = 2d3 1 + kuk2L2 ε−1 kA 2 uk2L2 + 2kf k2L∞ L2 .
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Navier friction boundary conditions
Applied Math Seminar, Sept. 17,24, 2008
One has
Z
t
Z
t
g (s)ds ≤ C ,
t−1
h(s)ds ≤ ε−1 k,
t−1
where k = k(κ) is small.
1
Note kA 2 u0 k2L2 ≤ C (ku0 k2H 1 + εδ−1 ku0 kL2 ) ≤ k(κ)ε−1 .
1
1
1
As far as (1 − 2d1 ε 2 kA 2 ukL2 ) ≥ 12 , equivalently, kA 2 uk2L2 ≤ dε−1 , one
1
estimates kA 2 u(t)k2L2 for t ≤ 1 by (usual) Gronwall’s inequality,
uses Uniform Gronwall’s inequality for t ≥ 1 to obtain
Z t
Z
Z t
1
1
2
2
kA 2 u(t)kL2 ≤
kA 2 u(s)kL2 ds +
h(s)ds exp
t−1
t−1
t
g (s)ds ,
t−1
The result is:
1
kA 2 u(t)k2L2 ≤ ε−1 k(κ).
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Navier friction boundary conditions
Applied Math Seminar, Sept. 17,24, 2008
THANK YOU FOR YOUR ATTENTION.
L. Hoang- Texas Tech
Navier friction boundary conditions
Applied Math Seminar, Sept. 17,24, 2008