On some singular limits for an atmosphere flow

Transcription

1 On some singular limits for an atmosphere flow Donatella Donatelli Dipartimento di Ingegneria e Scienze dell Informazione e Matematica Università degli Studi dell Aquila L Aquila, Italy donatella.donatelli@univaq.it

2 The model t u ε + (u ε ) u ε + 1 ε (g uε ) + 1 ε 2β pε = µ u ε 1 2 (div uε )u ε ε 2β t p ε + div u ε = 0, x Ω R 3, t 0, g = (0, 0, 1), β 0 2

3 The model t u ε + (u ε ) u ε + 1 ε (g uε ) + 1 ε 2β pε = µ u ε 1 2 (div uε )u ε ε 2β t p ε + div u ε = 0, x Ω R 3, t 0, g = (0, 0, 1), β 0 as ε 0? 2

4 Motivations it is a simplified model for some physical phenomena: atmospheric flow it is a simple model which includes the main feature of singular limits in fluid dynamics 3

5 Atmospheric flow model Compressible flow equations including rotation and gravity t u + u u + ɛg u + 1 ɛ 3 π = Q u t π + u π + γπ div u = Q π, many different scales: length scales from 10 5 m to 10 5 m, timescales from microseconds to weeks or more the Mach number is almost 0: the speed of the air at large scale is very very small in theoretical metereology is very important to understand the associated scale-specific phenomena (cloud formation, hurricanes, interacting waves at the tropics etc...) 4

6 A general scale transformation u ref =the thermal wind velocity h sc =density scale height c int =internal wave speed ɛ = u ref /c int l ref = h sc t ref = h sc /u ref We are interested in solutions with: T = t ref /ɛ αt the characteristic time, with α t > 0, L = l ref /ɛ αx the characteristic length α x > 0, Rescaled coordinates: t = ɛ αt t and x = ɛ αx x 5

7 Scaled model ɛ αt ɛ tu + u u + g u αx ɛ αx 1 + ɛαπ ɛ 3 π = Q u where ɛ αt ɛ t π + u π + γπ αx ɛ (div u) = Q π, απ π(t, x) = π(x) + ɛ απ Γ π(t, x). 6

8 Scaled model ɛ αt ɛ tu + u u + g u αx ɛ αx 1 + ɛαπ ɛ 3 π = Q u where ɛ αt ɛ t π + u π + γπ αx ɛ (div u) = Q π, απ π(t, x) = π(x) + ɛ απ Γ π(t, x). ɛ αt ɛ 1 αx Strouhal number ɛ αx 1 Rossby number 6

9 Advection time scale: α t = α x 7

10 Advection time scale: α t = α x Strouhal number = ɛ αt ɛ 1 αx Strouhal number 1 l ref t ref u ref = l ref = t ref u ref 7

11 Advection time scale: α t = α x Strouhal number = ɛ αt ɛ 1 αx Strouhal number 1 l ref t ref u ref = l ref = t ref u ref t u + u u + g u ɛ αx ɛ 3 απ π = Q u t π + u π + γπ ɛ απ (div u) = Q π, 7

12 Advection time scale: α t = α x Case A: α x 1 < 3 α π, 0 α π 3, α π 2 8

13 Advection time scale: α t = α x Case A: α x 1 < 3 α π, 0 α π 3, α π 2 t u + u u + g u ɛ αx ɛ 3 απ π = Q u t π + u π + γπ ɛ απ (div u) = Q π, The low Mach number regime dominates = 8

14 Advection time scale: α t = α x Case A: α x 1 < 3 α π, 0 α π 3, α π 2 t u + u u + g u ɛ αx ɛ 3 απ π = Q u t π + u π + γπ ɛ απ (div u) = Q π, The low Mach number regime dominates = incompressible fluid t u + u u u = π div u = 0 atmosphere models for the mesoscales 8

15 Advection time scale: α t = α x Case B: α x = α π = 2 9

16 Advection time scale: α t = α x Case B: α x = α π = 2 t u + u u + g u ɛ αx ɛ 3 απ π = Q u t π + u π + γπ ɛ απ (div u) = Q π, 9

17 Advection time scale: α t = α x Case B: α x = α π = 2 t u + u u + g u ɛ + 1 ɛ π = Q u t π + u π + γπ ɛ 2 (div u) = Q π, The speed of rotation and the incompressibility act on the same scale (we balance the Coriolis and the pressure gradient terms) = 9

18 Advection time scale: α t = α x Case B: α x = α π = 2 t u + u u + g u ɛ + 1 ɛ π = Q u t π + u π + γπ ɛ 2 (div u) = Q π, The speed of rotation and the incompressibility act on the same scale (we balance the Coriolis and the pressure gradient terms) = g u + π = 0, div u = 0. 9

19 Advection time scale: α t = α x Case B: α x = α π = 2 t u + u u + g u ɛ + 1 ɛ π = Q u t π + u π + γπ ɛ 2 (div u) = Q π, The speed of rotation and the incompressibility act on the same scale (we balance the Coriolis and the pressure gradient terms) = g u + π = 0, div u = 0. synoptic scale - quasi geostrophic approximation (geostrophic balance) 9

20 Our model t u ε + (u ε ) u ε + 1 ε (g uε ) + 1 ε 2β pε = µ u ε 1 2 (div uε )u ε ε 2β t p ε + div u ε = 0, 10

21 Our model t u ε + (u ε ) u ε + 1 ε (g uε ) + 1 ε 2β pε = µ u ε 1 2 (div uε )u ε ε 2β t p ε + div u ε = 0, Case A: α x 1 < 3 α π, 0 α π 3, α π 2: incompressible regime prevails β 1 10

22 Our model t u ε + (u ε ) u ε + 1 ε (g uε ) + 1 ε 2β pε = µ u ε 1 2 (div uε )u ε ε 2β t p ε + div u ε = 0, Case A: α x 1 < 3 α π, 0 α π 3, α π 2: incompressible regime prevails β 1 Case B: α x = α π = 2 geostrophic balance β = 1 2 g u + π = 0, div u = 0. 10

23 Our model t u ε + (u ε ) u ε + 1 ε (g uε ) + 1 ε 2β pε = µ u ε 1 2 (div uε )u ε ε 2β t p ε + div u ε = 0, Case A: α x 1 < 3 α π, 0 α π 3, α π 2: incompressible regime prevails β 1 Case B: α x = α π = 2 geostrophic balance β = 1 2 g u + π = 0, div u = 0. t ( h π π) + h π h( h π) = 2 h π. 10

24 In both cases we end up with a planar fluid! β 1 t u ε + (u ε ) u ε + 1 ε (g uε ) + 1 ε 2β pε = µ u ε 1 2 (div uε )u ε ε 2β t p ε + div u ε = 0, 11

25 In both cases we end up with a planar fluid! β 1 t u ε + (u ε ) u ε + 1 ε (g uε ) + 1 ε 2β pε = µ u ε 1 2 (div uε )u ε ε 2β t p ε + div u ε = 0, Apply the Leray projector H on the divergence free vector field space, as ε 0, we have H(g u) = 0, 11

26 In both cases we end up with a planar fluid! β 1 t u ε + (u ε ) u ε + 1 ε (g uε ) + 1 ε 2β pε = µ u ε 1 2 (div uε )u ε ε 2β t p ε + div u ε = 0, Apply the Leray projector H on the divergence free vector field space, as ε 0, we have H(g u) = 0, = g u = G, for a certain potential G 11

27 In both cases we end up with a planar fluid! β 1 t u ε + (u ε ) u ε + 1 ε (g uε ) + 1 ε 2β pε = µ u ε 1 2 (div uε )u ε ε 2β t p ε + div u ε = 0, Apply the Leray projector H on the divergence free vector field space, as ε 0, we have H(g u) = 0, = g u = G, for a certain potential G u = (u h, u 3 ) = (u 1, u 2, u 3 ) x h = (x 1, x 2 ) 11

28 In both cases we end up with a planar fluid! β 1 t u ε + (u ε ) u ε + 1 ε (g uε ) + 1 ε 2β pε = µ u ε 1 2 (div uε )u ε ε 2β t p ε + div u ε = 0, Apply the Leray projector H on the divergence free vector field space, as ε 0, we have H(g u) = 0, = g u = G, for a certain potential G u = (u h, u 3 ) = (u 1, u 2, u 3 ) x h = (x 1, x 2 ) ( u 2, u 1, 0) = G 11

29 In both cases we end up with a planar fluid! β 1 t u ε + (u ε ) u ε + 1 ε (g uε ) + 1 ε 2β pε = µ u ε 1 2 (div uε )u ε ε 2β t p ε + div u ε = 0, Apply the Leray projector H on the divergence free vector field space, as ε 0, we have H(g u) = 0, = g u = G, for a certain potential G u = (u h, u 3 ) = (u 1, u 2, u 3 ) x h = (x 1, x 2 ) ( u 2, u 1, 0) = G = x3 G = 0 11

30 In both cases we end up with a planar fluid! β 1 t u ε + (u ε ) u ε + 1 ε (g uε ) + 1 ε 2β pε = µ u ε 1 2 (div uε )u ε ε 2β t p ε + div u ε = 0, Apply the Leray projector H on the divergence free vector field space, as ε 0, we have H(g u) = 0, = g u = G, for a certain potential G u = (u h, u 3 ) = (u 1, u 2, u 3 ) x h = (x 1, x 2 ) ( u 2, u 1, 0) = G = x3 G = 0 = u h = u h (x h ) div u = 0 = x3 u 3 = 0, 11

31 In both cases we end up with a planar fluid! β 1 t u ε + (u ε ) u ε + 1 ε (g uε ) + 1 ε 2β pε = µ u ε 1 2 (div uε )u ε ε 2β t p ε + div u ε = 0, Apply the Leray projector H on the divergence free vector field space, as ε 0, we have H(g u) = 0, = g u = G, for a certain potential G u = (u h, u 3 ) = (u 1, u 2, u 3 ) x h = (x 1, x 2 ) ( u 2, u 1, 0) = G = x3 G = 0 = u h = u h (x h ) div u = 0 = x3 u 3 = 0, bd conditions = 11

32 In both cases we end up with a planar fluid! β 1 t u ε + (u ε ) u ε + 1 ε (g uε ) + 1 ε 2β pε = µ u ε 1 2 (div uε )u ε ε 2β t p ε + div u ε = 0, Apply the Leray projector H on the divergence free vector field space, as ε 0, we have H(g u) = 0, = g u = G, for a certain potential G u = (u h, u 3 ) = (u 1, u 2, u 3 ) x h = (x 1, x 2 ) ( u 2, u 1, 0) = G = x3 G = 0 = u h = u h (x h ) div u = 0 = x3 u 3 = 0, bd conditions = u 3 = 0 u = (u h (t, x h ), 0), u 3 = 0 11

33 In both cases we end up with a planar fluid! β = 1 2 t u ε + (u ε ) u ε + 1 ε (g uε ) + 1 ε pε = µ u ε 1 2 (div uε )u ε ε t p ε + div u ε = 0, 12

34 In both cases we end up with a planar fluid! β = 1 2 t u ε + (u ε ) u ε + 1 ε (g uε ) + 1 ε pε = µ u ε 1 2 (div uε )u ε ε t p ε + div u ε = 0, As ε 0, we have g u = π, div u = 0 + boundary conditions u = (u h (t, x h ), 0), u 3 = 0 12

35 A more technical remark Rotating incompressible fluid t u ε + (u ε ) u ε + 1 ε (g uε ) + 1 ε 2β pε = µ u ε div u ε = 0, 13

36 A more technical remark Rotating incompressible fluid t u ε + (u ε ) u ε + 1 ε (g uε ) + 1 ε 2β pε = µ u ε div u ε = 0, Artificial compressibility approximation t u ε + (u ε ) u ε + 1 ε (g uε ) + 1 ε 2β pε = µ u ε 1 2 (div uε )u ε ε t p ε + div u ε = 0, 13

37 A more technical remark Rotating incompressible fluid t u ε + (u ε ) u ε + 1 ε (g uε ) + 1 ε 2β pε = µ u ε div u ε = 0, Artificial compressibility approximation t u ε + (u ε ) u ε + 1 ε (g uε ) + 1 ε 2β pε = µ u ε 1 2 (div uε )u ε non increasing kinetic energy constraint ε 2β t p ε + div u ε = 0, linearized compressibility constraint 13

38 A more technical remark Rotating incompressible fluid t u ε + (u ε ) u ε + 1 ε (g uε ) + 1 ε 2β pε = µ u ε div u ε = 0, Artificial compressibility approximation t u ε + (u ε ) u ε + 1 ε (g uε ) + 1 ε 2β pε = µ u ε 1 2 (div uε )u ε non increasing kinetic energy constraint ε 2β t p ε + div u ε = 0, linearized compressibility constraint ρ t + div(ρu) = 0 equilibrium state: ρ = ρ 0, p = p(ρ 0 ) = p 0 linearizing... ρ t + ρ 0 div u = 0 for small perturbations: p p 0 = c 2 (ρ ρ 0 ) c = p (ρ 0 ) = sound speed 1 c 2 p t + ρ 0 div u = 0, 1 c 2 M 13

39 The model t u ε + 1 ε (g uε ) + 1 ε 2β pε = µ u ε (u ε ) u ε 1 2 (div uε )u ε ε 2β t p ε + div u ε = 0, x Ω R 3, t 0, g = (0, 0, 1), β 0 Ω = an infinite slab = R 2 (0, 1). Boundary conditions: u ε n Ω = 0, [Sn] n Ω = 0 Initial Conditions: u ε (x, 0) = u ε 0(x), p ε (x, 0) = p ε 0(x) u ε 0, p ε 0 L 2 (Ω) u ε 0 u 0, p ε 0 p 0 weakly in L 2 (R 3 ). we reformulate the problem in a periodic setting Ω = R 2 T 1, f(x 1, x 2, x 3 ) = f(x 1, x 2, x 3 ) 14

40 Main Result β 1 Assume that u ε, p ε are weak solutions of the system with initial and boundary data as before, then there exists u L 2 t Wx 1,2, such that u ε u weakly in L 2 (0, T ; W 1,2 (Ω)), u ε u strongly in L 2 loc ((0, T ) Ω), where u = [u h (t, x h ), 0] is the unique solution of the 2D incompressible Navier Stokes equation div h u = 0, t u + (u h )u + h π = h u. 15

41 Main Result β = 1/2 Assume that u ε, p ε are weak solutions of the system with initial and boundary data as before, then there exists u L 2 t Wx 1,2, π L t L 2 xsuch that u ε u u ε u p ε π where u and π satisfy weakly in L 2 (0, T ; W 1,2 (Ω)), strongly in L 2 loc ((0, T ) Ω), weakly in L (0, T ; L 2 (Ω)), div h u = 0, g u + π = 0 and π is a solution in the sense of distribution of the equation t ( h π π) + h π h( h π) = 2 h π 16

42 Energy estimate and uniform bounds E(t) = 1 2 E(t) + µ ( u ε (x, t) 2 + p ε (x, t) 2) dx. Ω t 0 Ω u ε (x, s) 2 dxds E(0). 17

43 Energy estimate and uniform bounds E(t) = 1 2 E(t) + µ ( u ε (x, t) 2 + p ε (x, t) 2) dx. Ω t 0 Ω u ε (x, s) 2 dxds E(0). p ε bd. in L t L 2 x, u ε bd. in L t L 2 x L 2 t L 6 x, u ε bd. in L 2 t,x, (u ε )u ε, (div u ε )u ε bd. in L 2 xl 1 x L 1 t L 3/2 x 17

44 Consequences εp ε 0 strongly in L 2 ([0, T ] Ω) u ε u weakly in L 2 ([0, T ]; H 1 (Ω)) div u = 0 a.e. in (0, T ) Ω H(g u) = 0 u = (u h (t, x h ), 0), u 3 = 0, u h = (u 1, u 2 ), x h = (x 1, x 2 ) 18

45 Consequences εp ε 0 strongly in L 2 ([0, T ] Ω) u ε u weakly in L 2 ([0, T ]; H 1 (Ω)) div u = 0 a.e. in (0, T ) Ω H(g u) = 0 u = (u h (t, x h ), 0), u 3 = 0, u h = (u 1, u 2 ), x h = (x 1, x 2 ) 19

46 Main problem: fast oscillating acoustic waves β 1 20

47 Main problem: fast oscillating acoustic waves β 1 t u ε + (u ε ) u ε + 1 ε (g uε ) + 1 ε 2β pε = µ u ε 1 2 (div uε )u ε ε 2β t p ε + div u ε = 0, 20

48 Main problem: fast oscillating acoustic waves β 1 ε 2β t u ε + p ε = ε 2β (µ u ε 1 2 (div uε )u ε (u ε ) u ε ) ε 2β 1 (g u ε ) ε 2β t p ε + div u ε = 0, 20

49 Main problem: fast oscillating acoustic waves β 1 ε 2β t u ε + p ε = ε 2β (µ u ε 1 2 (div uε )u ε (u ε ) u ε ) ε 2β 1 (g u ε ) ε 2β t p ε + div u ε = 0, Where do we see these waves? u ε = H[u ε ] }{{} + H [u ε ] }{{} soleinodal part gradient part div H[u ε ] = 0 H [u ε ] = Ψ u ε t + 1 ε 2β pε = 0 p ε t + 1 ε 2β div uε = 0 t Ψ + 1 ε 2β pε = 0 t p ε + 1 ε 2β Ψ = 0 20

50 Main problem: fast oscillating acoustic waves β = 1/2 21

51 Main problem: fast oscillating acoustic waves β = 1/2 t u ε + (u ε ) u ε + 1 ε (g uε ) + 1 ε pε = µ u ε 1 2 (div uε )u ε ε t p ε + div u ε = 0, 21

52 Main problem: fast oscillating acoustic waves β = 1/2 t u ε + 1 ε (g uε + p ε ) = µ u ε 1 2 (div uε )u ε (u ε ) u ε t p ε + 1 ε div uε = 0, 21

53 Main problem: fast oscillating acoustic waves β = 1/2 t u ε + 1 ε (g uε + p ε ) = µ u ε 1 2 (div uε )u ε (u ε ) u ε t p ε + 1 ε div uε = 0, Acoustic propagator: V = [ ] p u A [ ] ( ) p div u = u g u + p V t + 1 ε A[V] = 0 ˆV t (ξ, t) + 1 ε a(ξ) ˆV(ξ, t) = 0 Â[V](ξ) = a(ξ) ˆV(ξ) ˆV(ξ, t) = e t ε a(ξ) ˆV0 (ξ) 21

54 Acoustic waves β 1 ε 2β t p ε + div u ε = 0 ε 2β t u ε + p ε = ε 2β 1 F ε 1 + ε2β div F ε 2 + ε2β F ε 3, F ε 1 L t L 2 x, F ε 2 L 2 t L 2 x, F ε 3 L 2 t L 1 x 22

55 Acoustic waves β 1 ε 2β t p ε + div u ε = 0 ε 2β t u ε + p ε = ε 2β 1 F ε 1 + ε2β div F ε 2 + ε2β F ε 3, F ε 1 L t L 2 x, F ε 2 L 2 t L 2 x, F ε 3 L 2 t L 1 x Regularize the system: v δ = j δ v 22

56 Acoustic waves β 1 ε 2β t p ε + div u ε = 0 ε 2β t u ε + p ε = ε 2β 1 F ε 1 + ε2β div F ε 2 + ε2β F ε 3, F ε 1 L t L 2 x, F ε 2 L 2 t L 2 x, F ε 3 L 2 t L 1 x Regularize the system: v δ = j δ v ε 2β t p ε,δ + div u ε,δ = 0 ε 2β t u ε,δ + p ε,δ = ε 2β 1 F ε,δ 1 + ε 2β div F ε,δ 2 + ε 2β F ε,δ 3, F ε,δ 1 L 2 t Hk+ Fε,δ 2 L 2 t Hk+ Fε,δ 3 L 2 t Hk c(k, δ), for any k = 0, 1,... 22

57 Acoustic waves β 1 ε 2β t p ε + div u ε = 0 ε 2β t u ε + p ε = ε 2β 1 F ε 1 + ε2β div F ε 2 + ε2β F ε 3, F ε 1 L t L 2 x, F ε 2 L 2 t L 2 x, F ε 3 L 2 t L 1 x Regularize the system: v δ = j δ v ε 2β t p ε,δ + div u ε,δ = 0 ε 2β t u ε,δ + p ε,δ = ε 2β 1 F ε,δ 1 + ε 2β div F ε,δ 2 + ε 2β F ε,δ 3, F ε,δ 1 L 2 t Hk+ Fε,δ 2 L 2 t Hk+ Fε,δ 3 L 2 t Hk c(k, δ), for any k = 0, 1,... u ε,δ = Z ε,δ + Ψ ε,δ, where Z ε,δ = H[u ε,δ ], Ψ ε,δ = H [u ε,δ ] 22

58 Acoustic System β 1 ε 2β t p ε,δ + Ψ ε,δ = 0 ε 2β t Ψ ε,δ + p ε,δ = ε 2β 1 1 div F ε,δ 1 + ε 2β 1 div(div F ε,δ div F ε,δ 3 ) p ε,δ (0, ) = p ε,δ 0 Ψ ε,δ (0, ) = Q(u ε,δ 0 ) 23

59 Acoustic potential Ψ ε,δ (t) = 1 [ 2 ei t ε 2β Ψ ε,δ (0) + i ] p ε,δ (0) + 1 [ 2 e i t ε 2β Ψ ε,δ (0) i ] p ε,δ (0) t + ε t 0 t 0 (e i t s ε 2β +e i t s ε 2β )[ 1 div F ε,δ 1 ]ds (e i t s ε 2β +e i t s ε 2β )[ 1 divdiv F ε,δ 2 ]ds (e i t s ε 2β +e i t s ε 2β )[ 1 div F ε,δ 3 ]ds. 24

60 Estimates for the acoustic potential Dispersive estimates (D Ancona, Racke ARMA 2012) On any compact set K Ω, m > 0, we have and T 0 K T 0 t 0 K ( exp i t ) [v] ε m ( exp i t s ) ε m [g(s)] 2 dxdt ε m c v 2 L 2 (Ω), 2 dxdt ε m g L 2 ((0,T ) Ω). 25

61 Acoustic potential Ψ ε,δ (t) = 1 [ 2 ei t ε 2β Ψ ε,δ (0) + i ] p ε,δ (0) + 1 [ 2 e i t ε 2β Ψ ε,δ (0) i ] p ε,δ (0) t + ε t 0 t 0 (e i t s ε 2β +e i t s ε 2β )[ 1 div F ε,δ 1 ]ds (e i t s ε 2β +e i t s ε 2β )[ 1 divdiv F ε,δ 2 ]ds (e i t s ε 2β +e i t s ε 2β )[ 1 div F ε,δ 3 ]ds. 26

62 Estimates for the acoustic potential Dispersive estimates (D Ancona, Racke ARMA 2012) On any compact set K Ω, m > 0, we have and T 0 K T 0 t 0 K ( exp i t ) [v] ε m ( exp i t s ) ε m [g(s)] 2 dxdt ε m c v 2 L 2 (Ω), 2 dxdt ε m g L 2 ((0,T ) Ω). Decay estimate for the acoustic potential T 0 for any compact set K Ω Ψ ε,δ 2 L 2 (K) dt (ε2β 1 + ε 2β )c(δ, K, T ) 27

63 Convergence β 1-step 1 28

64 Convergence β 1-step 1 T 0 Ψ ε,δ 2 L 2 (K) dt (ε2β 1 + ε 2β )c(δ, K, T ) as ε

65 Convergence β 1-step 1 T 0 Ψ ε,δ 2 L 2 (K) dt (ε2β 1 + ε 2β )c(δ, K, T ) as ε 0 0 u ε,δ = Z ε,δ + Ψ ε,δ, where Z ε,δ = H[u ε,δ ], Ψ ε,δ = H [u ε,δ ] 28

66 Convergence β 1-step 1 T 0 Ψ ε,δ 2 L 2 (K) dt (ε2β 1 + ε 2β )c(δ, K, T ) as ε 0 0 u ε,δ = Z ε,δ + Ψ ε,δ, where Z ε,δ = H[u ε,δ ], Ψ ε,δ = H [u ε,δ ] (u ε,δ )u ε,δ uε,δ div u ε,δ = div(z ε,δ Z ε,δ ) div( Ψ ε,δ Ψ ε,δ )+div(z ε,δ Ψ ε,δ )+div( Ψ ε,δ Z ε,δ ) 1 2 uε,δ Ψ ε,δ 28

67 Convergence β 1-step 1 T 0 Ψ ε,δ 2 L 2 (K) dt (ε2β 1 + ε 2β )c(δ, K, T ) as ε 0 0 u ε,δ = Z ε,δ + Ψ ε,δ, where Z ε,δ = H[u ε,δ ], Ψ ε,δ = H [u ε,δ ] (u ε,δ )u ε,δ uε,δ div u ε,δ = div(z ε,δ Z ε,δ ) + div( Ψ ε,δ Ψ ε,δ )+div(z ε,δ Ψ ε,δ )+div( Ψ ε,δ Z ε,δ ) 1 2 uε,δ Ψ ε,δ 28

68 Convergence β 1-step 2 div(z ε,δ Z ε,δ ) = 1 2 Zε,δ 2 Z ε,δ curl[z ε,δ ] Establish the compactness of the vertical average of Z ε,δ Z ε,δ (x h ) = 1 T 1 T 1 Z ε,δ (x h, x 3 )dx 3 29

69 Convergence β 1-step 2 div(z ε,δ Z ε,δ ) = 1 2 Zε,δ 2 Z ε,δ curl[z ε,δ ] Establish the compactness of the vertical average of Z ε,δ Z ε,δ (x h ) = 1 T 1 T 1 Z ε,δ (x h, x 3 )dx 3 Z ε,δ u δ, strongly in L 2 ((0, T ) K) 29

70 Convergence β 1-step 2 div(z ε,δ Z ε,δ ) = 1 2 Zε,δ 2 Z ε,δ curl[z ε,δ ] Establish the compactness of the vertical average of Z ε,δ Z ε,δ (x h ) = 1 T 1 T 1 Z ε,δ (x h, x 3 )dx 3 Z ε,δ u δ, strongly in L 2 ((0, T ) K) we can infer that the oscillations are due to the vector fields that depends on x 3 29

71 Convergence β 1-step 2 div(z ε,δ Z ε,δ ) = 1 2 Zε,δ 2 Z ε,δ curl[z ε,δ ] Establish the compactness of the vertical average of Z ε,δ Z ε,δ (x h ) = 1 T 1 T 1 Z ε,δ (x h, x 3 )dx 3 Z ε,δ u δ, strongly in L 2 ((0, T ) K) we can infer that the oscillations are due to the vector fields that depends on x 3 we estimate the oscillations {Z ε,δ }(x) = Z ε,δ (x) Z ε,δ (x h ). and we show that they don t interfere in the convergence of the nonlinear terms. 29

72 Acoustic waves β = 1/2 ε t p ε + div u ε = 0 ε t u ε + (g u ε + p ε ) = ε div G ε 1 + εgε 2, G ε 1 L 2 t L 2 x, G ε 2 L 2 t L 1 x 30

73 Acoustic waves β = 1/2 ε t p ε + div u ε = 0 ε t u ε + (g u ε + p ε ) = ε div G ε 1 + εgε 2, G ε 1 L 2 t L 2 x, G ε 2 L 2 t L 1 x Weak formulation: T (εp ε t ϕ + u ε ϕ) dxdt = ε p ε 0ϕ(0, )dx. T 0 Ω ( εu ε t ϕ (g u ε ) ϕ + p ε div ϕ ) dxdt = Ω 0 Ω T ε (G ε 1 : ϕ + G ε 2 ϕ)dx ε u ε 0ϕ(0, )dx, 0 Ω Ω 30

74 Acoustic propagator A : L 2 (Ω) L 2 (Ω; R 3 ) R 4 [ ] ( ) p div u A = u g u + p 31

75 Acoustic propagator A : L 2 (Ω) L 2 (Ω; R 3 ) R 4 [ ] ( ) p div u A = u g u + p Goal? 31

76 Acoustic propagator A : L 2 (Ω) L 2 (Ω; R 3 ) R 4 [ ] ( ) p div u A = u g u + p Goal? to prove that the component of the vector (p ε, u ε ) orthogonal to the null space of A decays to zero as ε 0 31

77 Acoustic propagator A [ ] ( ) p div u = u g u + p 32

78 Acoustic propagator [ ] ( ) p div u A = u g u + p { } Ker(A) = p = p(x h ), u = u(x h ), div h u h = 0, h p = (u 2, u 1 ). 32

79 Acoustic propagator [ ] ( ) p div u A = u g u + p { } Ker(A) = p = p(x h ), u = u(x h ), div h u h = 0, h p = (u 2, u 1 ). eigenvalues problem for A in the frequency space is set as follows: ( 2 i ξ j û j +kû 3 ) λˆp = 0, i(ξ 1, ξ 2, k)ˆp (û 2, û 1, 0) λû = 0. j=1 λ 2 = 1 + ξ 2 + k 2 ± (1 + ξ 2 + k 2 ) 2 4k

80 Acoustic propagator [ ] ( ) p div u A = u g u + p { } Ker(A) = p = p(x h ), u = u(x h ), div h u h = 0, h p = (u 2, u 1 ). eigenvalues problem for A in the frequency space is set as follows: ( 2 i ξ j û j +kû 3 ) λˆp = 0, i(ξ 1, ξ 2, k)ˆp (û 2, û 1, 0) λû = 0. j=1 λ 2 = 1 + ξ 2 + k 2 ± (1 + ξ 2 + k 2 ) 2 4k 2. 2 the only real eigenvalue is λ = 0 for k = 0 = the space of eigenvectors of A coincides with the Ker(A) 32

81 The RAGE theorem (Cycon at al, Cycon et al., Schrödinger operators: with applications to quantum mechanics and global geometry, 1987, Theorem 5.8) Let H be a Hilbert space, A : D(A) H H a self-adjoint operator, C : H H a compact operator, and P c the orthogonal projection onto H c, specifically, } H = H c cl H {span{w H w an eigenvector of A}. Then 1 τ Moreover τ 0 exp( ita)cp c exp(ita) dt 0 as τ. L(H) 33

82 The RAGE theorem (Cycon at al, Cycon et al., Schrödinger operators: with applications to quantum mechanics and global geometry, 1987, Theorem 5.8) Let H be a Hilbert space, A : D(A) H H a self-adjoint operator, C : H H a compact operator, and P c the orthogonal projection onto H c, specifically, } H = H c cl H {span{w H w an eigenvector of A}. Then 1 τ Moreover 1 T 2 τ 0 exp( ita)cp c exp(ita) dt 0 as τ. L(H) 1 T t C 0 T 0 t C exp(i ε A)P cx, 2 A)X(s)ds exp(i t s ε where ω(ε) 0, as ε 0. 2 H L 2 (0,T ;H) dt ω(ε) X 2 H, T dt ω(ε) X(s) 2 Hds. 0 33

83 Application of the RAGE theorem H = H M = {(p, u) ˆp(ξ h, k) = 0, û(ξ h, k) = 0 if ξ h + k > M} A = ia, C[v] = P M [χv], χ C c (Ω), 0 χ 1, where P M : L 2 (Ω) L 2 (Ω; R 3 ) H M is the orthogonal projection into H M. ( ) p ε M u ε M = exp(ia t ( ) p ε ) ε M (0) u ε M (0) + t 0 exp(ia t s ( ) 0 ε ) ds G ε,m 34

84 Q is the orthogonal projection into Ker(A), Q : L 2 (Ω) L 2 (Ω; R 3 ) Ker(A) 35

85 Q is the orthogonal projection into Ker(A), Q : L 2 (Ω) L 2 (Ω; R 3 ) Ker(A) the point spectrum of A, is reduced to 0 35

86 Q is the orthogonal projection into Ker(A), Q : L 2 (Ω) L 2 (Ω; R 3 ) Ker(A) the point spectrum of A, is reduced to 0 ( ) Q p ε M 0 in L 2 ((0, T ) K; R 4 ), as ε 0, u ε M ( ) p ε Q M u ε M ( πm u M for any compact set K Ω and fixed M. ) in L 2 ((0, T ) K; R 4 ), as ε 0, 35

87 Convergence P M u ε P M u in L 2 ((0, T ) K; R 3 ), 36

88 Convergence P M u ε P M u in L 2 ((0, T ) K; R 3 ), u ε u weakly in L 2 ([0, T ]; H 1 (Ω)), 36

89 Convergence P M u ε P M u in L 2 ((0, T ) K; R 3 ), u ε u weakly in L 2 ([0, T ]; H 1 (Ω)), u ε u in L 2 ((0, T ) K; R 3 ), for any compact set K Ω 36

90 Convergence-final step T 0 Ω for any ψ C c ([0, T ) Ω) ( p ε t ψ + 1 ) ε uε ψ dxdt = p ε 0ψ(0, )dx, Ω T 0 Ω ( u ε t ϕ + u ε u ε : ϕ 1 2 uε div u ε ϕ + 1 ) ε (g uε ) ϕ dxdt = T 0 Ω u ε ϕdxdt u ε 0 ϕ(0, )dx, Ω for any ϕ = ( h ψ, 0) = ( x 2 ψ, x1 ψ, 0), ψ C c ([0, T ) Ω). 37

91 T 0 Ω...as ε 0 ( ) u t h ψ + u u : h ψ + π tψ dxdt = T 0 Ω u h dxdt Ω u 0 h (0, ) + p 0ψ(0, )dx. 38

92 T 0 Ω...as ε 0 ( ) u t h ψ + u u : h ψ + π tψ dxdt = T 0 Ω u h dxdt g u + π = 0 u = h π Ω u 0 h (0, ) + p 0ψ(0, )dx. 38

93 T 0 T 0 Ω Ω...as ε 0 ( ) u t h ψ + u u : h ψ + π tψ dxdt = T 0 Ω u h dxdt g u + π = 0 u = h π Ω u 0 h (0, ) + p 0ψ(0, )dx. ( ) h π t h ψ + h π h π : h ψ + π tψ dx h dt = T 0 Ω h π h ψdx hdt (u 0 h ψ(0, ) + p 0ψ(0, ))dx Ω 38

94 T 0 T 0 Ω Ω...as ε 0 ( ) u t h ψ + u u : h ψ + π tψ dxdt = T 0 Ω u h dxdt g u + π = 0 u = h π Ω u 0 h (0, ) + p 0ψ(0, )dx. ( ) h π t h ψ + h π h π : h ψ + π tψ dx h dt = T 0 Ω h π h ψdx hdt (u 0 h ψ(0, ) + p 0ψ(0, ))dx Ω weak formulation of t ( h π π) + h π h( h π) = 2 h π. 38

95 To conclude... Figure from: R. Klein. Scale-dependent models for atmospheric flows. Annual. Rev. Fluid Mechanics, 42, ,