Lubrication and roughness

Transcription

1 Lubrication and roughness Laurent Chupin 1 & Sébastien Martin Institut Camille Jordan - Lyon 2 - Laboratoire de Mathématiques - Orsay GdR CHANT - August 2010 Laurent Chupin (ICJ Lyon) Lubrication and roughness GdR CHANT 1 / 19

2 Summary 1 Thin lms: lubrication 2 Roughness: wall laws 3 Lubrication and roughness Ansatz Stokes problems Reynolds problems Algorithm Error analysis Two particular cases Laurent Chupin (ICJ Lyon) Lubrication and roughness GdR CHANT 1 / 19

3 Summary 1 Thin lms: lubrication 2 Roughness: wall laws 3 Lubrication and roughness Ansatz Stokes problems Reynolds problems Algorithm Error analysis Two particular cases Laurent Chupin (ICJ Lyon) Lubrication and roughness GdR CHANT 1 / 19

4 Anisotropy and shear Anisotropic domain Ω ε = { } (x, y) R d+1 ; 0 < y < εh + (x) { (u, v) = (s, 0) on the bottom y = 0, Shear velocity (u, v) = (0, 0) on the top y = εh + (x). Laurent Chupin (ICJ Lyon) Lubrication and roughness GdR CHANT 2 / 19

5 Asymptotic behavior x u 2 y u + xp = 0 on Ω ε x v 2 y v + y p = 0 on Ω ε div x u + y v = 0 on Ω ε Denoting (u 0, v 0, p 0 ) the equivalent obtained when ε tends to 0, we obtain p 0 only depends on x div x ( h xp 0 ) = div x ( h 2 s ) on R d (u 0, v 0 ) are explicit fonctions of p 0 and Z = y ε References: Reynolds (1886) Bayada-Chambat (1986) Wilkening (2009) Laurent Chupin (ICJ Lyon) Lubrication and roughness GdR CHANT 3 / 19

6 Summary 1 Thin lms: lubrication 2 Roughness: wall laws 3 Lubrication and roughness Ansatz Stokes problems Reynolds problems Algorithm Error analysis Two particular cases Laurent Chupin (ICJ Lyon) Lubrication and roughness GdR CHANT 3 / 19

7 Roughness-induced eective boundary conditions u(x, y) = u 0 (x, y) + ε ũ 1 ( x ε, y ε ) + ε u 1 (x, y) + First order estimates u u 0 L2 (Ω ε) ε and u u 0 H 1 (Ω ε) ε. Next order: justication of the law u + εα y u = 0 on y = 0. References: Navier (1827) Jäger-Mikelic (2001) Gérard Varet (2008) Laurent Chupin (ICJ Lyon) Lubrication and roughness GdR CHANT 4 / 19

8 Summary 1 Thin lms: lubrication 2 Roughness: wall laws 3 Lubrication and roughness Ansatz Stokes problems Reynolds problems Algorithm Error analysis Two particular cases Laurent Chupin (ICJ Lyon) Lubrication and roughness GdR CHANT 4 / 19

9 Summary 1 Thin lms: lubrication 2 Roughness: wall laws 3 Lubrication and roughness Ansatz Stokes problems Reynolds problems Algorithm Error analysis Two particular cases Laurent Chupin (ICJ Lyon) Lubrication and roughness GdR CHANT 4 / 19

10 Proposition of development u(x, y) = v(x, y) = p(x, y) = N j=0 N j=0 N j=0 [ ( ε j u j x, y ) + ε ũ j+1 (x, x ε ε, y )] + R(x, y), 2 ε 2 [ ( ε j+1 v j x, y ) + ṽ j+1 (x, x ε ε, y )] + S(x, y), 2 ε 2 [ ( ε j 2 p j x, y ) + ε p j+1 (x, x ε ε, y )] + Q(x, y). 2 ε 2 Laurent Chupin (ICJ Lyon) Lubrication and roughness GdR CHANT 5 / 19

11 Proposition of development u(x, y) = v(x, y) = p(x, y) = N j=0 N j=0 N j=0 [ ( ε j u j x, y ) + ε ũ j+1 (x, x ε ε, y )] + R(x, y), 2 ε 2 [ ( ε j+1 v j x, y ) + ṽ j+1 (x, x ε ε, y )] + S(x, y), 2 ε 2 [ ( ε j 2 p j x, y ) + ε p j+1 (x, x ε ε, y )] + Q(x, y). 2 ε 2 1 Put this development in the Stokes system 2 Identify the same order with respect to ε 3 Separate the variables x, Z = y ε, X = x ε 2 and Y = y ε 2 4 Treat the boundary conditions Laurent Chupin (ICJ Lyon) Lubrication and roughness GdR CHANT 5 / 19

12 Systems satised at each order Stokes system for the unknowns (ũ j, ṽ j, p j ) on the cell ω bl : (S (j) ) Example of source term X ũ j 2 Y ũj + X p j = F j on ω bl, X ṽ j 2 Y ṽj + Y p j = G j on ω bl, div X ũ j + Y ṽ j = H j on ω bl, ũ j = Ũ j on γ bl, ṽ j = Ṽj on γ bl, (ũ j, ṽ j, p j ) is X periodic. H j : (X, Y ) C j 2 div x ũ j 2 (, X, Y ) Boundary terms Ũ j : X h (X) Z u j 1 (x, 0) Ṽ j : X h (X) Z v j 2 (x, 0) Laurent Chupin (ICJ Lyon) Lubrication and roughness GdR CHANT 6 / 19

13 Systems satised at each order Reynolds system for the unknowns (u j, v j, p j ) on the domain ω R : 2 Z u j + x p j = F j on ω R, Z p j = G j on ω R, div x u j + Z v j = H j on ω R, (u j, v j ) = (0, 0) on γ 0, (u j, v j ) = (α j, β j+1 ) on γ +. (R (j) ) Example of source term H j : x C j (x). Boundary term α j (x) = β j+1 (x) = lim Y + ]0,1[d ũj(x, X, Y ) d X lim Y + ]0,1[d ṽj+1(x, X, Y ) d X Laurent Chupin (ICJ Lyon) Lubrication and roughness GdR CHANT 7 / 19

14 Summary 1 Thin lms: lubrication 2 Roughness: wall laws 3 Lubrication and roughness Ansatz Stokes problems Reynolds problems Algorithm Error analysis Two particular cases Laurent Chupin (ICJ Lyon) Lubrication and roughness GdR CHANT 7 / 19

15 Analysis of the Stokes problems Proposition 1 There exists constants A j 2, B j 1 and C j 2 such that the solution of S (j) satises, for all X ]0, 1[ d and Y > 0 ũ j (X, Y ) α j and ṽ j (X, Y ) β j and p j (X, Y ) e Y Proof: 1 Induction for j 2 Fourier analysis Note that for j N, we have choosen C j = div x α j. Laurent Chupin (ICJ Lyon) Lubrication and roughness GdR CHANT 8 / 19

16 Analysis of the Stokes problems One important remark: Lemma 1 If we dene L v : ( ]0,1[d H, Ũ, Ṽ) ṽ(x, 0) d X R, where ṽ is a function X periodic satisfying { div X ũ + Y ṽ = H on ω bl, (ũ, ṽ) = (Ũ, Ṽ) on γ bl, then ( ) L v H, c h (X), c h (X) = {Y <0} H(X, Y ) d XdY c ]0,1[d h (X) d X. Laurent Chupin (ICJ Lyon) Lubrication and roughness GdR CHANT 9 / 19

17 Summary 1 Thin lms: lubrication 2 Roughness: wall laws 3 Lubrication and roughness Ansatz Stokes problems Reynolds problems Algorithm Error analysis Two particular cases Laurent Chupin (ICJ Lyon) Lubrication and roughness GdR CHANT 9 / 19

18 Analysis of the Reynolds problems Proposition 2 For j N the problem R (j) admits a unique regular solution (u j, v j, p j ). Proof - Due to the fact that C j = div x α j, the system R (j) implies that u j + α j, v j + β j+1 and p j verify the following Reynolds-type problems: 2 Z u + xp = F on ω R, Z p = G on ω R, div x u + Z v = 0 on ω R, (u, v) = (U 0, V 0 ) on γ 0, (u, v) = (0, 0) on γ +. where U 0 = α j and V 0 = β j+1. Compatibility condition: Td V0 (x) dx = 0. Laurent Chupin (ICJ Lyon) Lubrication and roughness GdR CHANT 10 / 19

19 Summary 1 Thin lms: lubrication 2 Roughness: wall laws 3 Lubrication and roughness Ansatz Stokes problems Reynolds problems Algorithm Error analysis Two particular cases Laurent Chupin (ICJ Lyon) Lubrication and roughness GdR CHANT 10 / 19

20 Algorithm 2 Z u 0 + x p 0 = 0 on ω R, Z p 0 = 0 on ω R, div x u 0 + Z v 0 = 0 on ω R, X ũ 1 2 Y ũ1 + X p 1 = 0 on ω bl, X ṽ 1 2 Y ṽ1 + Y p 1 = 0 on ω bl, div X ũ 1 + Y ṽ 1 = 0 on ω bl, (u 0, v 0 ) = (s, 0) on γ 0, (u 0, v 0 ) = (0, 0) on γ +. ũ 1 = h (X) Z u 0 (x, 0) on γ bl, ṽ 1 = 0 on γ bl. This corrector produce an error in the macroscopic domain Ω + ε α 1 (x) := lim Y + ]0,1[d ũ1(x, X, Y ) d X 0. 2 Z (u 1 + α 1 ) + x p 1 = 0 on ω R, Z p 1 = 0 on ω R, div x (u 1 + α 1 ) + Z v 1 = 0 on ω R, since (u 1 + α 1, v 1 ) = (α 1, 0) on γ 0, (u 1 + α 1, v 1 ) = (0, 0) on γ +. Laurent Chupin (ICJ Lyon) Lubrication and roughness GdR CHANT 11 / 19

21 Algorithm X ũ 2 2 Y ũ2 + X p 2 = 0, X ṽ 2 2 Y ṽ2 + Y p 2 = 0, div X ũ 2 + Y ṽ 2 = 0, ũ 2 = h (X) Z u 1 (x, 0) on γ bl, ṽ 2 = h (X) Z v 0 (x, 0) on γ bl. This corrector produce an error in the macroscopic domain Ω + ε α 2 (x) := lim Y + ]0,1[d ũ2(x, X, Y ) d X. since 2 Z (u 2 + α 2 ) + x p 2 = x u 0, Z p 2 = 2 Z v 0, div x (u 2 + α 2 ) + Z (v 2 + β 3 ) = 0, (u 2 + α 2, v 2 + β 3 ) = (α 2, β 3 ) on γ 0, (u 2 + α 2, v 2 + β 3 ) = (0, 0) on γ +. The function β 3 can be evaluate using the result of lemma 1: β 3 (x) = L v ( H 3, Z u 2 (x, 0) h, Z v 1 (x, 0) h ) ( ) = div x ũ 1 (x, X, Y ) d XdY. Y <0 Laurent Chupin (ICJ Lyon) Lubrication and roughness GdR CHANT 12 / 19

22 Summary 1 Thin lms: lubrication 2 Roughness: wall laws 3 Lubrication and roughness Ansatz Stokes problems Reynolds problems Algorithm Error analysis Two particular cases Laurent Chupin (ICJ Lyon) Lubrication and roughness GdR CHANT 12 / 19

23 Stokes system satised by the residue x R 2 y R + x Q = F ε (N) on Ω ε, x S 2 y S + y Q = G ε (N) on Ω ε, div x R + y S = H ε (N) on Ω ε, R = U ε (N) on Γ + ε, S = V ε (N) R = 0 on Γ + ε, on Γ ε, S = W ε (N) on Γ ε. Example of source term: ( F (N) ε (x, y) = F R ε x, y ) + F bl ε (x, x ε ε, y ) 2 ε 2 with F R ε :=ε N 1( ) ε x u N + x u N 1 and F bl ε :=ε N 2( ε 3 x ũ N+1 + ε 2 x ũ N + ε x ũ N 1 + x ũ N 2 + 2ε x X ũ N x X ũ N ε x p N+1 x p N ). Laurent Chupin (ICJ Lyon) Lubrication and roughness GdR CHANT 13 / 19

24 Lift velocity: Bogovskii formulae 1 Lift velocity for the boundary condition (easy and explicit) 2 Lift velocity for the divergence condition: Proposition 3 (Bogovskii) If H Ḣ m (Ω ε ) then there exists a solution ( R, S) H m+1 (Ω ε ) of { div x R + y S = H on Ωε, ( R, S) = (0, 0) on Γ ε Γ + ε, such that x,y ( R, S) H m (Ωε) 1 ε H Hm (Ωε). Laurent Chupin (ICJ Lyon) Lubrication and roughness GdR CHANT 14 / 19

25 Classical estimates The unknowns (R := R R, S := S S, Q) satisfy x R 2 y R + x Q = F ε on Ω ε, x S 2 y S + y Q = G ε on Ω ε, div x R + y S = 0 on Ω ε, (R, S) = (0, 0) on Γ ε Γ + ε. Proposition 4 i) Estimates in the L 2 norm: (R, S) L2 (Ω ε) ε 2 (F ε, G ε ) L2 (Ω ε), Q L2 (Ω ε) ε (F ε, G ε ) L2 (Ω ε). ii) Estimates in the H 1 norm: (R, S) H 1 (Ω ε) ε (F ε, G ε ) L2 (Ω ε), Q H 1 (Ω ε) (F ε, G ε ) L2 (Ω ε). Laurent Chupin (ICJ Lyon) Lubrication and roughness GdR CHANT 15 / 19

26 Control with respect to ε Lemma 2 Let f C 1 (T d ω bl ) such that f (,, Y ) = O(e Y ) for Y +. ( f x, x ε, y ) 2 ε 2 L 2 (Ω ε and f ε) (x, x ε, y ) 2 ε 2 H 1 (Ω 1 ε) ε. Lemma 3 Let g C 0 (ω R ) dened on {(x, Z), Z < 0} by a regular extension. ( g x, y ) ε L 2 (Ω ε) ε. Using the denition of the source terms F ε, G ε and H ε, we obtain F ε L2 (Ω ε) ε N 1, G ε L2 (Ω ε) ε N 3/2, H ε H 1 (Ω ε) ε N 1... Laurent Chupin (ICJ Lyon) Lubrication and roughness GdR CHANT 16 / 19

27 Control with respect to ε Proposition 5 i) Estimates in the L 2 norm: (R, S) L2 (Ω ε) ε N 1/2, Q L2 (Ω ε) ε N 3/2. ii) Estimates in the H 1 norm: (R, S) H 1 (Ω ε) ε N 3/2, Q H 1 (Ω ε) ε N 5/2. Laurent Chupin (ICJ Lyon) Lubrication and roughness GdR CHANT 17 / 19

28 Summary 1 Thin lms: lubrication 2 Roughness: wall laws 3 Lubrication and roughness Ansatz Stokes problems Reynolds problems Algorithm Error analysis Two particular cases Laurent Chupin (ICJ Lyon) Lubrication and roughness GdR CHANT 17 / 19

29 Contribution of the rugosities in the thin lm approximation Without rugosities: h = 0 Then the asymptotic expansion reduces to u (x, y) = v (x, y) = p (x, y) = [N/2] j=0 [N/2] j=0 [N/2] j=0 ( ε 2j u 2j x, y ) + R(x, y), ε ( ε 2j+1 v 2j x, y ) + S(x, y), ε ( ε 2j 2 p 2j x, y ) + Q(x, y). ε Only the sequence of Reynolds problems R (2j) is considered. Laurent Chupin (ICJ Lyon) Lubrication and roughness GdR CHANT 18 / 19

30 Scale eects due to the curvature of the lm thickness Without curvature: h + = constant 1 (u 0, v 0, p 0 ) explicit and simple (in particular x u 0 = 0 and v 0 = 0). 2 (ũ 1, ṽ 1, p 1 ) solution of S (1) (does not depend on x). 3 For j > 0, we obtain (case d = 1) u j + α j = α j ũ j = α j s ũ1, s u 0, v j = 0, p j = α j s p 0 ṽ j = α j s ṽ1, p j = α j s p 1 and α j s ( = α1 ) j. s 4 We deduce the following approximate solution (error e h+ /ε ) s u = s + (u 0 s) + ε α 1 ũ 1, s ε α 1 s ε α 1 v = ε α 1 ṽ 1, s ε α 1 p = 1 s ε 2 p α 1 p 1. s ε α 1 ε s ε α 1 Laurent Chupin (ICJ Lyon) Lubrication and roughness GdR CHANT 19 / 19