Navier-Stokes equations in thin domains with Navier friction boundary conditions

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1 Navier-Stokes equations in thin domains with Navier friction boundary conditions Luan Thach Hoang Department of Mathematics and Statistics, Texas Tech University lhoang/ Applied Mathematics Seminar Texas Tech University September 17, 4, 008

2 Outline 1 Introduction Main results 3 Functional settings 4 Linear and non-linear estimates 5 Global solutions

3 Introduction Navier-Stokes equations for fluid dynamics: t u + (u )u ν u = p + f, div u = 0, u(x, 0) = u 0 (x), ν > 0 is the kinematic viscosity, u = (u 1, u, u 3 ) is the unknown velocity field, p R is the unknown pressure, f (t) is the body force, u 0 is the given initial data.

4 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) = 1 ( u + ( u) ).

5 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 x 3 = const, then the conditions become the Robin conditions (see [Hu]) u 3 = 0, u 1 + γ 3 u 1 = u + γ 3 u = 0. Compressible fluids on half planes with Navier friction boundary conditions [Hoff]. Assume ν = 1.

6 Thin domains Ω = Ω ε = {(x 1, x, x 3 ) : (x 1, x ) T, h0(x ε 1, x ) < x 3 < h1(x ε 1, x )}, where ε (0, 1], h0 ε = εg 0, h1 ε = εg 1, and g 0, g 1 are given C 3 functions defined on T, g = g 1 g 0 c 0 > 0. The boundary is Γ = Γ 0 Γ 1, where Γ 0 is the bottom and Γ 1 is the top.

7 Boundary conditions on thin domains The velocity u satisfies the Navier friction boundary conditions on Γ 1 and Γ 0 with friction coefficients γ 1 = γ ε 1 and γ 0 = γ ε 0, respectively. Assumption There is δ [0, 1], such that for i=0,1, 0 < lim inf ε 0 γi ε γi ε lim sup εδ ε 0 ε δ <.

8 Notation Leray-Helmholtz decomposition L (Ω ε ) 3 = H H where H = {u L (Ω ε ) 3 : u = 0 in Ω ε, u N = 0 on Γ}, H = { φ : φ H 1 (Ω ε )}.

9 Notation Leray-Helmholtz decomposition where L (Ω ε ) 3 = H H H = {u L (Ω ε ) 3 : u = 0 in Ω ε, u N = 0 on Γ}, H = { φ : φ H 1 (Ω ε )}. Let V be the closure in H 1 (Ω ε, R 3 ) of u C (Ω ε, R 3 ) H that satisfies the friction boundary conditions.

10 Notation Leray-Helmholtz decomposition where L (Ω ε ) 3 = H H H = {u L (Ω ε ) 3 : u = 0 in Ω ε, u N = 0 on Γ}, H = { φ : φ H 1 (Ω ε )}. Let V be the closure in H 1 (Ω ε, R 3 ) of u C (Ω ε, R 3 ) H that satisfies the friction boundary conditions. Averaging operator: M 0 φ(x ) = 1 h1 φ(x, x 3 )dx 3, Mu = (M0 u 1, M 0 u, 0). εg h 0

11 Main result Theorem (Global strong solutions) Let δ [/3, 1]. There are ε 0 > 0 and κ > 0 such that if ε (0, ε 0 ] and u 0 V and f L (L ) satisfy m u,0 = Mu 0 L, m u,1 = ε u 0 H 1, m f,0 = Mf L L, m f,1 = ε f L L, are smaller than κ, then the regular solution exists for all t 0: u C([0, ), H 1 (Ω ε )) L loc ([0, ), H (Ω ε )). Remark: The condition on u 0 is acceptable.

12 A Green s formula [Solonnikov-Šcǎdilov] u v dx = Ω [ (Du : Dv) + ( u)( v)] dx + {((Du)N) v ( u)(v N)} dσ. Ω Ω If u is divergence-free and satisfies the Navier friction boundary conditions, v is tangential to the boundary then u v dx = Ω ε (Du : Dv) dx + γ 0 Ω ε u v dσ + γ 1 Γ 0 u v dσ. Γ 1

13 A Green s formula [Solonnikov-Šcǎdilov] u v dx = Ω [ (Du : Dv) + ( u)( v)] dx + {((Du)N) v ( u)(v N)} dσ. Ω Ω If u is divergence-free and satisfies the Navier friction boundary conditions, v is tangential to the boundary then u v dx = Ω ε (Du : Dv) dx + γ 0 Ω ε u v dσ + γ 1 Γ 0 u v dσ. Γ 1 The right hand side is denoted by E(u, v).

14 Uniform Korn inequality Is E(, ) bounded and coercive in H 1 (Ω ε )? We need Korn s inequality: u H 1 (Ω ε) C εe(u, u). Lemma There is ε 0 > 0 such that for ε (0, ε 0 ], u H 1 (Ω ε ) H0 tangential to the boundary of Ω ε, one has and u is u L Cε 1 δ E(u, u), C u H 1 E(u, u) C ( u L + ε 1 δ u L ), where C, C are positive constants independent of ε.

15 Stokes operator Let P denotes the (Leray) projection on H. Then the Stokes operator is: Au = P u, u D A, D A = {u H (Ω ε ) 3 V : u satisfies the Navier friction boundary conditions} For u D A, v V, one has Navier-Stokes equations: Au, v = E(u, v). du + Au + B(u, u) = Pf, dt where B(u, v) = P(u u).

16 Inequalities For ε (0, ε 0 ], one has the following: If u V = D A 1 then u L Cε (1 δ)/ A 1 u L, u H 1 C A 1 u L, If u D A then A 1 u L C( u L + ε (δ 1)/ u L ). A 1 u L Cε (1 δ)/ Au L, u L Cε 1 δ Au L.

17 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, u N τ = u One also has [Chueshov-Raugel-Rekalo] { N τ γτ }. N ( u) = N {N (( N) u) γn u}. Our case: N ε and γ ε δ.

18 Linear and Non-linear Estimates Proposition If ε (0, ε 0 ] and u D A, then Au + u L C 1 ε δ u L + C 1 ε δ 1 u L, C Au L u H C 3 Au L.

19 Linear and Non-linear Estimates Proposition If ε (0, ε 0 ] and u D A, then Au + u L C 1 ε δ u L + C 1 ε δ 1 u L, C Au L u H C 3 Au L. Proposition There is ε 0 > 0 such that for ε (0, ε 0 ], α > 0 and u D A, one has u u, Au {α + Cε 1/ A 1 u L } Au L + C α ε δ u L A 1 u 4 L + C α ε 1 ε δ /3 u /3 L A 1 u L + C α ε 1 u L A 1 u L. where the positive number C α depends on α but not on ε.

20 Corollary Suppose δ [/3, 1], then there exists ε (0, 1] such that for any ε < ε and u D A, one has { 1 } u u, Au 4 + d 1ε 1/ A 1 u L Au L } + d { u L A 1 u L } A 1 u L + d 3 {1 + u L ε 1 A 1 u L. where positive constants d 1, d and d 3 are independent of ε.

21 Key identity Lemma Let u D A and Φ H 1 (Ω ε ) 3. One has ( ( u)) Φdx = ( Φ) ( u + G(u))dx Ω ε Ω ε Φ ( G(u))dx. Ω ε where G(u) Cε δ u, G(u) Cε δ u + Cε δ 1 u.

22 Key identity Lemma Let u D A and Φ H 1 (Ω ε ) 3. One has ( ( u)) Φdx = ( Φ) ( u + G(u))dx Ω ε Ω ε Φ ( G(u))dx. Ω ε where G(u) Cε δ u, G(u) Cε δ u + Cε δ 1 u. Linear estimate: Φ = Au + u, Φ = 0. Non-linear estimate: Φ = u ( u).

23 Estimate of u L Lemma There is ε 0 (0, 1] such that if ε < ε 0 and u H (Ω ε ) 3 satisfies the Navier friction boundary conditions, then u L C u L + C u H 1. Remarks on the proof. Integration by parts ( 1 u dx = u u dx + Ω ε Ω ε Γ N u ) N u dσ.

24 Estimate of u L Lemma There is ε 0 (0, 1] such that if ε < ε 0 and u H (Ω ε ) 3 satisfies the Navier friction boundary conditions, then u L C u L + C u H 1. Remarks on the proof. Integration by parts ( 1 u dx = u u dx + Ω ε Ω ε Γ N u ) N u dσ. Remove the second derivatives in the boundary integrals

25 Estimate of u L Lemma There is ε 0 (0, 1] such that if ε < ε 0 and u H (Ω ε ) 3 satisfies the Navier friction boundary conditions, then u L C u L + C u H 1. Remarks on the proof. Integration by parts ( 1 u dx = u u dx + Ω ε Ω ε Γ N u ) N u dσ. Remove the second derivatives in the boundary integrals Appropriate order for ε

26 Estimate of u L Lemma There is ε 0 (0, 1] such that if ε < ε 0 and u H (Ω ε ) 3 satisfies the Navier friction boundary conditions, then u L C u L + C u H 1. Remarks on the proof. Integration by parts ( 1 u dx = u u dx + Ω ε Ω ε Γ N u ) N u dσ. Remove the second derivatives in the boundary integrals Appropriate order for ε The role of the positivity of the friction coefficients

27 Estimate of u L Lemma There is ε 0 (0, 1] such that if ε < ε 0 and u H (Ω ε ) 3 satisfies the Navier friction boundary conditions, then u L C u L + C u H 1. Remarks on the proof. Integration by parts ( 1 u dx = u u dx + Ω ε Ω ε Γ N u ) N u dσ. Remove the second derivatives in the boundary integrals Appropriate order for ε The role of the positivity of the friction coefficients u L u L + C u H 1 + Cε u L.

28 Estimate of the non-linear term Write u = v + w where v = Mu = ( Mu, Mu ψ), ψ(x) = 1 εg {(x 3 h 0 ) h 1 + (h 1 x 3 ) h 0 }. Then v is divergence free and tangential to the boundary. Important properties: v is a D-like vector field. w satisfies good inequalites: v L 4 Cε 1/4 u 1/ u 1/, v L H 1 L 4 Cε 1/4 u 1/ u 1/ H 1 H Then write w L Cε w L, w L Cε u H + Cε δ u L, w L Cε 1/ u H + Cε δ/ u 1/ L u 1/ H. (u )u, Au = (w )u, Au + (v )u, Au + u (v )u, u.

29 Strong global solutions Steps: Do not need u = (v, w) and equations for each v and w Non-linear estimate and Uniform Gronwall s inequality Estimates for u(t) L Estimates for A 1 u(t) L and t t 1 A 1 u(s) L ds and the right ε 1 size.

30 L -Estimates for u Poincaré-like inequalities: (I M)u L Cε u H 1. 1 d dt u L + A 1 u L u, Pf ˆMu, ˆMPf + (I ˆM)u, (I ˆM)Pf 1 A 1 u L + MPf L L + ε Pf L L. Hence Gronwall s inequality yields u(t) L u 0 L e c1t + MPf L L + ε Pf L L. u 0 L = Mu 0 L + (I M)u 0 L Mu 0 L + Cε u 0 H 1 Cκ. u(t) L, t+1 t A 1 u L ds Cκ.

31 H 1 -Estimate for u 1 d { dt A 1 u 1 L + Au L 4 + d 1ε 1/ A 1 u L } Au L } +d { u L A 1 u L } A 1 u L +d 3 {1+ u L ε 1 A 1 u L + Au L f L L. where d dt A 1 u L + (1 d 1 ε 1 A 1 u L ) Au L g A 1 u L + h, g = d u L A 1 u L, h = d 3 { 1 + u L }ε 1 A 1 u L + f L L.

32 One has t t 1 g(s)ds C, t t 1 h(s)ds ε 1 k, where k = k(κ) is small. Note A 1 u 0 L C( u 0 H 1 + ε δ 1 u 0 L ) k(κ)ε 1. As far as (1 d 1 ε 1 A 1 u L ) 1, equivalently, A 1 u L dε 1, one estimates A 1 u(t) L for t 1 by (usual) Gronwall s inequality, uses Uniform Gronwall s inequality for t 1 to obtain ( t t ) A 1 u(t) L A 1 u(s) L ds + h(s)ds exp t 1 t 1 The result is: A 1 u(t) L ε 1 k(κ). ( t ) g(s)ds, t 1

33 THANK YOU FOR YOUR ATTENTION.