Asymptotic Analysis of Acoustic Waves in a Porous ... - NYU (Math)

Asymptotic Analysis of Acoustic Waves in a Porous Medium: Micro-Incompressible Flow Jose Diaz-Alban

∗

and Nader Masmoudi†

October 28, 2013

Key words: Acoustic waves, Compressible Navier-Stokes, Stokes, porous medium, Bloch decomposition, boundary layers. AMS subject classification: 35B27, 35B40, 76N20, 76M50. Abstract This is the second in a series of three papers which studies acoustic waves governed by the linearized compressible Stokes equations in a porous medium. In particular, we want to analyze the simultaneous inviscid and high frequency limits of fluid flows in a porous medium. The presence of time-space boundary layers decouples the flow into an incompressible (that we call micro-incompressible) and an acoustic part (that we call micro-acoustic) on the microscopic scale. While this paper employs the two-scale methods used in our first paper [10], the present boundary layer phenomenon requires additional weak convergence tools. Using the Bloch decomposition, we introduce modified Helmholtz operators, enabling us to split the flow into its micro-incompressible and micro-acoustic parts. Closed equations for the micro-incompressible flow are obtained using two-scale convergence, while closed equations for the micro-acoustic flow are given in our forthcoming paper.

1

Introduction

The homogenization of the Stokes operator in a porous medium is well studied. We refer the interested reader to text books [3, 24, 16] for some formal developments and to [25, 1, 20] for some rigorous mathematical results. However, there are fewer works dealing with the homogenization of the acoustic system or of the compressible Navier-Stokes system (see for instance [7, 19, 11, 18, 29]). Here, we continue the study of the homogenization of the compressible Navier-Stokes system linearized around (1, 0) in a porous medium which was started in [10]. ∗

[email protected] , [email protected], Courant Institute of Mathematical Sciences. Partially supported by NSF grant DMS-1211806 †

1

The linearized compressible Navier-Stokes equations in a porous medium read 8 ∂t pε + divuε = 0 > > > > > ε β ε ε > > < ∂t u − ε 4u + ∇p = 0 uε = 0

> > > > > > > :

ε

p (t = 0) = b

(0, ∞) × Rdε (1)

∂ Rdε

ε

uε (t = 0) = aε

where pε and uε are the pressure and velocity of the fluid, and εβ is the viscosity. Here, Rdε is the porous medium formed by periodic repetition of an elementary fluid cell which has been shrunk to size ε (see section 2.1). Finally, 1 < β measures the relative importance of the viscous effect to the size of each cell. We refer the reader to [10] for a complete study of the case of strongly viscous flow (β ≤ 1) and for a derivation of this model. In the case of 1 < β, the presence of time-space boundary layers decouples the flow into an incompressible (that we call micro-incompressible, see Definition 3.5) and an acoustic part (that we call micro-acoustic, see Definition 3.5) on the microscopic scale. We remark that this is in contrast to the case of β ≤ 1, where not only is the boundary layer in space absent, but the boundary layers in time decouples the velocity and pressure when ε goes to zero (see [10]). In order to motivate the goals of this paper, we begin with a formal approach to the case when β = 2 (we refer the interested reader to [24] and [19] for a formal and rigorous study when β=2). Formal asymptotic expansions expressing (pε , uε ) in the variables (t, x, y), y = xε , yields the limiting terms (p0 , u0 ) and the associated two-scale system Z 8 > > |Yf |∂t p0 + divx u0 = 0 > > > Yf > > > > > ∂t u0 − 4y u + ∇x p0 + ∇y p1 = 0 > > > > > > divy u0 = 0 > > > > > ∇y p 0 = 0 > < u0 = 0 > Z > > > > p1 = 0 > > > Yf > > > > > > p0 (t = 0) = b(x) > > > > > u > 0 (t = 0) = Py a0 (x, y) > > : y → (p1 , u0 ) is Y -periodic

in Rd in Rd × Yf

on ∂Ys

(2)

Here p1 is the Lagrange multiplier corresponding to the incompressibility of u0 and Py is the divergence free Helmholtz operator on the domain Yf (see Remark 3.1). We notice that two-scale convergence captures the micro-incompressible part of the flow. What is not so apparent is the formation of a thin boundary layer in time of size ε1/2 which traps the energy carried by the micro-acoustic part of the flow when ε goes to zero. Throughout this paper we identify the micro-acoustic part of the flow as (q ε , v ε ). The overall behavior of the flow near initial time can be seen in the following diagram.

2

(bε , aε )

(q ε , v ε )

(pε − q ε , uε − v ε )

t

ε1/2

Figure 1: Behavior of the flow near t = 0 when β = 2. We remark that the time behavior associated the micro-acoustic part of the flow is non-trivial. In fact, the initial layer in time of size ε1/2 and time oscillations associated to (q ε , v ε ) are generated by the presence of a boundary layer in space of size ε3/2 . This paper is devoted to the complete study of the micro-incompressible flow for 1 < β. Our goal is to describe the energy carried by the micro-incompressible part of the flow as ε goes to zero. Mathematically, we wish to understand the local and total energies eε (t, x) = |pε (t, x) − q ε (t, x)|2 + |uε (t, x) − v ε (t, x)|2 (3)

E ε (t) = kpε (t) − q ε (t)k2L2 (Rd ) + kuε (t) − v ε (t)k2L2 (Rd ) ε

ε

when fluid flow occurs in a porous medium and the limit ε going to zero is taken. We note that the energy carried by the micro-acoustic part of the flow is subject of our final paper [9]. A complete understanding of the underlying time-space boundary layer phenomena is necessary in order to propose appropriate asymptotic expansions describing the energy carried by micro-incompressible flow. In the case 1 < β < 2, we show that the boundary layer in time of size ε2−β decouples the microincompressible part of the flow when ε goes to zero. More precisely, the usual formal two-scale expansions in the variables (t, x, y), y = xε , will describe the pressure pε − q ε , but will fail to describe the velocity uε − v ε for all times. In order to describe the velocity near initial time, we need to introduce the time scale τ = t/ε2−β during the homogenization process. Through this time rescaling, we deduce that the velocity associated to micro-incompressible flow is governed by the Stokes system (16) near initial time. The overall behavior of the flow near initial time can be seen in the following diagram.

(bε , aε )

(q ε , v ε )

uε − v ε

ε(3−β)/2

3

pε − q ε

ε2−β

t

Figure 2: The behavior of the flow near t=0 for 1 < β < 2. We make a few observations. First, we note that the boundary layer in time of size ε2−β remains from the strongly viscous regime (see [10]). Secondly, the time behavior that impacts the micro-acoustic part of the flow, namely the boundary layer in time of size ε(3−β)/2 and associated time oscillations are non-trivial and are generated by a boundary layer in space of size ε(β+1)/2 . This time behavior associated to micro-acoustic flow is essentially given in [6] (see also [15] for a more precise construction) for the case β = 2, and we refer the reader to [8], [9] for further details in the general case 1 < β. Lastly, we invite the reader to compare with Figure 1 and note that boundary layer in time of size ε2−β breaks when β = 2, hence the coupling of the pressure and velocity of micro-incompressible flow. When 2 < β, there is no boundary layer in time that impacts the micro-incompressible part of the flow. Nevertheless, using a formal approach to yield the asymptotic behavior of micro-incompressible flow yields the following information “ x” = u0 (τ, x, y) 6= 0 lim (uε − v ε ) t, x, ε→0+ ε

on ∂Ys .

The failure of the no-slip condition on the microscopic level is due to the formation of a boundary layer in space of size εβ/2 , which forms as ε vanishes. It follows that linearized Prandtl boundary layer expansions are necessary for the case 2 < β, and we hope that the reader will find the presentation given here interesting in its own right. Using the Prandtl expansions, we show that this layer quickly dissipates the energy carried by the waves. That is, the energy carried by micro-incompressible flow is described by the closed equations obtained via the usual two-scale process. In closing, we leave the reader with a few remarks concerning micro-incompressible flow. First, we note that while β = 2 is in a sense a critical value, we do not assign definitions to the regimes 1 < β < 2 and 2 < β. Also, while there is a boundary layer in space of size ε(β+1)/2 when 1 < β, we only briefly mentioned it since its impact is mainly related to micro-acoustic flow, which is the subject of our forthcoming paper [9]. Moreover, while this paper employs the two-scale methods used in our first paper, the present boundary layer phenomena requires additional weak convergence tools. For instance, as we mentioned earlier the Prandtl expansions are necessary for the case 2 < β, but also incorporated in the homogenization process is the Bloch decomposition. We refer the reader to [28], [5] and [12] as examples of just some of works involving the Bloch decomposition. Using the Bloch decomposition we introduce ε , Qε (see section 3.1), which in essence serve as the basis for the the modified Helmholtz operators PM M current article as well as our forthcoming work [9]. Indeed, it is through these modified operators that we are able to partition the flow into its incompressible and acoustic parts on the microscopic level. What’s more, we believe the modified Helmholtz operators to be new and interesting in their own right.

2 2.1

Preliminaries The Domain

Define Y = (0, 1)d to be the unit open cube in Rd , and let Ys be a closed smooth subset of Y with strictly positive measure. By smoothness of ∂Ys , we mean to take it as regular as needed. The domain Y /Ys is denoted by Yf and we refer to Ys , Yf to be the solid and fluid parts of Y . Repeating Yf by Y -periodicity to all of Rd we get the fluid domain R0 . The porous medium is now defined to be εR0 which is denoted by Rdε . We define Yfε , Ysε to be εYf , εYs and note Rdε = {x : x ∈ Yfε + p,

4

for some p ∈ Zd }.

Figure 3: The elementary fluid cell

2.2

Notations

We define B = [0, 2π)d and we say that a function f (x) defined on Rd0 is θ-quasi periodic, for θ ∈ B, provided f (x + p) = eiθ·p f (x), ∀p ∈ Zd . (4) A function which is 0-quasi periodic, is simply referred to as Y -periodic. The space of smooth functions which vanish at infinity is denoted by D(Rd ), and for θ ∈ B, Cθk (Yf ) is the space of differentiable functions up to order k which are θ-quasi periodic. We define the space L2θ (Yf ) to be all functions in L2loc (Rd0 ) satisfying (4). The Sobolev space Hθk (Yf ) is the space of the functions k (Y is the subspace of functions in with derivatives up to order k belonging to L2θ (Yf ) and the space H0θ f Hθk (Yf ) that vanish on the boundary ∂Ysε . Also recall that the space Hθ−1 (Yf ) is defined by the norm kf kH −1 (Y θ

f)

=

sup

|< f, u >|

kukH 1 (Y ) =1 f 0θ

k (Y ), L2 (Y ), H k (Y ), H k (Y ), H −1 (Y ). We denote X d to be When θ = 0, we use the notations C# f f # f # f 0# f # the Banach space of vector valued functions of d-components, each component belonging to Banach space X. We note that all the Hilbert spaces mentioned in this paper are equipped with an inner product whose second component is conjugated, here z ∗ denotes the complex conjugate of z. Lastly, the we define the operators [ · ]1 , [ · ]2 acting on Cd+1 as

[z]1 = z1 ,

[z]2 = z2 ,

where z = (z1 , z2 ) ∈ C × Cd .

2.3

Two-Scale Convergence

Two-scale convergence is a notion first introduced by G. Nguetseng, [21], and later extended by G. Allaire [2], used to capture the parts of f ε which oscillate at frequency ε−1 . More precisely, two-scale convergence is a rigorous justification of the first term in formal two-scale expansions. In our problem two-scale convergence plays a significant role in understanding the local and total energies (3). Definition 2.1 Let f ε (x), f0 (x, y) belong to L2 (Rdε ), L2 (Rd × Y ) respectively. We say f ε (x) two-scale 2−s

converges to f0 (x, y) (denoted f ε * f0 ) provided Z Z Z “ x” ε→0 f ε (x) σ x, → f0 (x, y) σ(x, y), ε Rd Rd Yf ε ∞ (Y )). for all σ ∈ D(Rd ; C#

5

Notice that if f ε (x) two-scale converges to f0 (x, y) then the following always holds lim kf ε k2L2 (Rd ) ≥ kf0 k2L2 (Rd ×Y

ε→0

f)

ε

.

If equality holds, we have 2−s

Definition 2.2 We say f ε strongly two-scale converges to f0 (x, y), denoted f ε → f0 (x, y) provided f ε two-scale converges to f0 and lim kf ε k2L2 (Rd ) = kf0 k2L2 (Rd ×Y ) . ε→0

ε

Strong two-scale convergence is interpreted as saying that all of the oscillations of f ε are of frequency ε−1 . This is due to the fact that strong two-scale convergence implies ‚ “ x ”‚ ‚ ‚ (5) lim kf ε kL2 (Rd ) = lim ‚f0 x, ‚ ε ε→0 ε→0 ε L2 (Rdε ) whenever f0 is continuous in either the x or y variable. More generally, (5) holds provided f0 is an admissible test function in the sense of Allaire (see definition 1.4 of [2]). We note that given ` f0´(x, y), having continuity in at least one of its variables enables us to make sense of the quantity f0 x, xε . We refer the reader to Section 5 of [2] for further discussion and to [10] for a characterization of strong two-scale convergence. If f0 (x, y) is an admissible test function, then we can improve the convergence of (5). Indeed, we have the following Z ˛ “ x ”˛2 ˛ ε ˛ lim ˛f (x) − f0 x, ˛ = 0. ε→0 Rd ε ε Finally, we can add a time dependence in all the limits we mentioned above. In particular if f0 ∈ C([0, T ]; L2# (Yf ; C(Rd ) ∩ L2 (Rd ))), then we have ‚ “ x ”‚ ‚ ‚ lim ‚f0 t, x, = kf0 kL2 ((0,T )×Rd ×Yf ) . ‚ ε→0 ε L2 ((0,T )×Rdε ) The main result of two-scale convergence is the existence of a two-scale limit for bounded sequences in L2 (Rdε ). The main result is stated below; for a proof we refer the reader to [21], [2]. Theorem 2.3 Let f ε belong to a bounded set of L2 (Rdε ). Then there exist a subsequence of ε (still denoted by ε) and a Y -periodic function f0 (x, y) ∈ L2 (Rd × Y ), such that f ε (x) two-scale converges to f0 (x, y).

2.4

The Bloch Decomposition of A0

In order to completely state the main results of this paper, we introduce the Bloch decomposition of the wave operator „ « 0 div A0 = , ∇ 0 on the domain Yfε . Here, ε will be fixed and one can think as if ε = 1. We keep it in the notation for later use. We begin by defining Z L2⊕,ε = − L2θ (Yfε ) dθ B

to be the space of all measurable mappings u : B → L2θ (Yfε ) with finite norm kuk2L2

⊕,ε

Z Z =− −

|u(x, θ)|2 dx dθ < ∞.

B Yfε

Due to the presence of the holes in our problem, one uses the test functions φ ∈ Cc∞ (Rdε ) to deduce the following proposition. The proof is essentially given in [23].

6

Proposition 2.4 The mapping I ε : L2 (Rdε ) → L2⊕,ε defined as X

(I ε u)(x, θ) = u# (x, θ) =

u(x − εp) eiθ·p ,

p∈Zd

is unitary. The inverse is given by Z ` ε −1 ´ (I ) u# (x) = − u# (x, θ) dθ. B

If u, v belong to L2 (Rdε ), then Parseval’s formula holds, (u, v)L2 (Rd ) = (u# , v# )L2 ε

⊕,ε

.

We now study the spectral problem with Neumann boundary conditions. More precisely, we define for each θ ∈ B, the non-decreasing eigenvalues {µ2k (θ)}1≤k and eigenvectors { φk (x, θ)}1≤k satisfying 8 > > >
∂ν > > : φk (x + p, θ) = ei θ·p φk (x, θ),

on ∂Ys

(6)

∀p ∈ Zd

The following proposition summarizes some of the basic properties of the eigenvalues and eigenfunctions µk (θ), φk (θ). A proof is provided in subsection 4.1. Proposition 2.5 For each θ ∈ [0, 2π)d there exist µk (θ), φk (θ), k = 1, 2, 3 . . . satisfying (6) such that k→∞

0 ≤ µ21 (θ) ≤ µ22 (θ) ≤ . . . ≤ µ2k (θ) . . . → ∞. Moreover: (i) µk (θ) belongs to C# (B) and satisfies sup θ,ξ

|µk (θ) − µk (ξ)| ≤1 |θ − ξ|

θ6=ξ

k→∞

(ii) 0 = inf µ1 (θ) < inf µ2 (θ) ≤ . . . inf µk (θ) ≤ . . . → ∞. Moreover, µ1 (0) = 0. θ∈B

θ∈B

θ∈B

(iii) The principle eigenvalue µ1 (θ) is simple in a neighborhood of the origin. (iv) For all θ ∈ B, the eigenvectors {φk (θ)}1≤k form an orthonormal basis of L2θ (Yf ). For θ 6= 0, the n o ∇φk (θ) sequence forms an orthonormal basis of Hθ (Yf ) where µ (θ) k

1≤k

Hθ (Yf ) = {u ∈ L2θ (Yf )d : u = ∇p for some p ∈ Hθ1 (Yf )}.

7

(v) When θ = 0, the sequences {φk (0)}1 : vk (x + p, θ) = vk (x, θ), p ∈ Zd

in Yf on ∂Ys

The diagonalization of the inviscid operator A0 on the domain Yfε is now obtained using the eigenvalues and eigenfunctions of the spectral problem (6) along with an appropriate rescaling. Indeed, for each θ 6= 0 and k ∈ Zd /{0}, we define λεk (θ) = ε−1 λk (θ), “x ” ,θ , Φεk (x, θ) = ε−d/2 Φk ε

(7)

where λk (θ) = sgn(k)µ|k| (θ), 0 φ (x,θ) |k|

√ 2

B Φk (x, θ) = B @

sgn(k)∇φ|k| (x,θ) √ i µ|k| (θ) 2

1 C C. A

Remark 2.7 When θ = 0, we use the same definition (7), but take 1 < |k|. We conclude

8 > > < > > :

A0 Φεk (θ) = iλεk (θ)Φεk (θ)

in Yfε

[Φεk (θ)]2 · ν = 0 Φεk (x

i θ·p

+ p, θ) = e

on ∂Ysε Φεk (x, θ),

(8)

d

p∈Z .

Now, for each θ ∈ B, we define Skε (θ) to be the space spanned by Φεk (θ) and we define Πεk (θ) to be the orthogonal projection from L2θ (Yfε )d+1 onto Skε (θ). We define the following mutually orthogonal subspaces on L2 (Rdε )d+1 Z Skε = (I ε )−1 − Skε (θ) dθ B

= {u ∈ L2 (Rdε )d+1 ; u# (·, θ) ∈ Skε (θ) for a.e. θ ∈ B},

8

where I ε is the mapping given in proposition 2.4. We denote Πεk as the orthogonal projection from L2 (Rdε )d+1 onto Skε . Notice that we have Z Πεk Φ = (I ε )−1 − Πεk (θ)I ε Φ dθ . B

As a result of Proposition 2.5, the mutually orthogonal spaces Skε (θ) have the following direct sum decomposition in L2θ (Yfε )d+1 : M

Skε (θ) = L2θ (Yfε ) × Hθ (Yfε ),

when θ 6= 0.

(9)

1≤|k|

Integrating in θ and applying (I ε )−1 to (9), we have the following result. The proof will be given in subsection 4.2. Proposition 2.8 M

Skε ⊆ L2 (Rdε ) × H(Rdε ) ⊆

Skε = L2 (Rdε ) × Hloc (Rdε ),

(10)

1≤k

1 < > > :

in Yfε

−4g = divf

on ∂Ysε

∇g(θ) · ν = f · ν i θ·p

g(x + εp, θ) = e

g(x, θ),

9

d

∀p ∈ Z

(13)

It is not difficult to see that P ε f, Qε f indeed yield the usual Helmholtz operators. That is, (

div(P ε f ) = 0 P εf · ν = 0

on Rdε on ∂Rdε

and that Qε f is a gradient (see [8]). What we find interesting is how Helmholtz operators P ε , Qε defined in (12) can also be characterized through the projection operators Πεk . Indeed, as a consequence of direct sum (9), we have the following Qε (θ)f (θ) =

X

[ Πεk (θ)(0, f (θ)) ]2 ,

when θ 6= 0.

1≤|k|

Here f (θ) belongs to L2θ (Yfε )d and (0, f (θ)) ∈ C × Cd . It follows that Qε f =

X

[ Πεk (0, f ) ]2

1≤|k|

for f ∈ L2 (Rdε )d . Remark 3.1 When θ = 0, we denote Pyε , Qεy the Helmholtz operators on the domain Yfε . That is Pyε = I − Qεy where Qεy f = ∇g and g is defined by the system 8 −4g = divf in Yfε > > > > > > ·ν =f ·ν on ∂Ysε > < ∇g Z g=0 > > > > Yfε > > > : g is Y -periodic

(14)

As a result of Proposition 2.5, we have the following direct sum in L2# (Yfε )d+1 M

Skε (0) = L2# (Yfε )/R × H# (Yfε ),

1 > > div u = 0 y > > > > > u=0 on ∂Ys > < Z > > > > > > > > > > > :

p=0

(16)

Yf

u(τ = 0) = Py a0 (x, y) y → (p, u) is Y -periodic

Remark 3.7 The relationship between the time variables τ and t is given by τ = t/ε2−β . This is due to the initial layer in time of size ε2−β that remains from the case of strongly viscous flow (see [10]). Notice that this layer in time decouples the micro-incompressible part of the flow as ε → 0+ (see Figure 2). We also note that if (bε2 , QεM aε ) goes strongly to 0 in L2 , then (q ε , v ε ) can be replaced by 0 in the statement of the theorem. Theorem 3.8 (β = 2)Let (S1) hold, and assume T ∈ (0, ∞). Assume sufficient regularity on the initial data (b(x), Py a0 (x, y)) and let (uε , pε ) be the solution to (1) for β = 2. Then ε→0

sup kpε (t, x) − p(t, x) − q ε (t, x)k2L2 (Rd ) → 0, ε

0≤t≤T

‚2 ‚ “ ε→0 x” ‚ ‚ − v ε (t, x)‚ → 0, sup ‚uε (t, x) − u t, x, 2 d ε L (Rε ) 0≤t≤T ‚2 Z T‚ “ ” ‚ ‚ ε→0 x 1 ‚∇uε − ∇y u t, x, − ∇v ε (t, x)‚ ε2 ‚ 2 d → 0, ‚ ε ε 0

L (Rε )

where (p(t, x), u(t, x, y)) is the solution to the following two-scale Stokes/Acoustic system Z 8 > |Yf |∂t p + divx u=0 in Rd > > > Yf > > > > > > in Rd × Yf > > ∂t u − 4 y u + ∇ x p + ∇ y p 1 = 0 > > > divy u = 0 > > > > < u=0 on ∂Ys Z > > > p1 = 0 > > > Yf > > > > > p(t = 0) = b(x) > > > > > u(t = 0) = Py a0 (x, y) > > > : y → (p1 , u) is Y -periodic

12

(17)

Remark 3.9 Notice that the layer in time of size ε2−β disappears at β = 2, hence the micro-incompressible part of the flow becomes coupled. In particular, we obtain the coupled macro-micro system given in (17). We refer the reader to Figure 1 for the diagram illustrating the behavior of the flow near initial time. Theorem 3.10 (2 < β) Let (S1) hold, and assume T ∈ (0, ∞). Assume sufficient regularity on the initial data (b(x), Py a0 (x, y)) and let (uε , pε ) be the solution to (1) for 2 < β. Then ε→0

sup kpε (t, x) − p (t, x) − q ε (t, x)k2L2 (Rd ) → 0, ε

0≤t≤T

‚ ‚2 “ ε→0 x” ‚ ‚ sup ‚uε (t, x) − u t, x, − v ε (t, x)‚ → 0, 2 d ε L (R ) 0≤t≤T ε Z T ε→0 ε ε 2 β k∇u − ∇v kL2 (Rd ) → 0, ε ε

0

where (p(t, x), u(t, x, y)) is the solution to the following two-scale acoustic system Z 8 > u=0 in Rd |Yf |∂t p + divx > > > Yf > > > > > > ∂t u + ∇ x p + ∇ y p 1 = 0 in Rd × Yf > > > > > divy u = 0 > > > > < u·ν =0 on ∂Ys Z > > > p1 = 0 > > > Yf > > > > > p(t = 0) = b(x) > > > > > u(t = 0) = Py a0 (x, y) > > > : y → (p1 , u) is Y -periodic

(18)

Remark 3.11 For β > 2, while an initial layer in time is not present, we do have a boundary layer in space of size εβ/2 due to the fact that we can only impose the boundary condition u · ν = 0 on ∂Ys . A description of the boundary layer flow in space is necessary to conclude the overall behavior of the fluid. We refer the reader to the proof of Theorem 3.10 where we apply linearized Prandtl boundary layer expansions to conclude that the energy trapped is dissipated by heat flow. We conclude this section with a few observations regarding the convergence results of this section. Remark 3.12 (i) All the energy descriptions we obtained for 1 < β have a macroscopic behavior, contrasting with the behavior of the flow when β ≤ 1. We refer the reader to [10], where there is no macroscopic flow due to very strong viscosity. (ii) Notice the intensity in which energy is dissipated weakens as β increases. More precisely, notice that the micro-incompressible part of the flow only contributes to the dissipation of the energy provided β ≤ 2. We show in our final paper [9] that the micro-acoustic part of the flow never dissipates the energy when 3 < β. That is εβ

Z

ε→0

|∇v ε |2 → 0 Rd ε

13

whenever 3 < β

(iii) The oscillations of the pressure term pε are completely contained in q ε whenever β > 1. (iv) In the three convergence Theorems for 1 < β, we did not try to give a rate of convergence in terms of ε. One can easily get some error estimate in terms of ε by computing the next terms in the formal asymptotic expansions, but this is not our goal in this paper. (v) Also, all our Theorems for 1 < β have the assumption that the initial condition (b(x), Py a0 (x, y)) is sufficiently regular. We did not try to get the best possible regularity on the initial data and one can easily lower the regularity requirement on the initial data by performing a mollification in the x variable. We refer the reader to [10] for details.

4

Proof of modified Helmholtz decomposition

The main goal of this section is to prove Theorem 3.4. We start with the proof of some propositions

4.1

Proof of proposition 2.5:

For θ ∈ B, we recall the following characterization

µ2k (θ) =

min

max

F ∈Σk (θ) u∈F

k∇uk2L2 (Y

f)

kuk2L2 (Y

,

(19)

f)

where Σk (θ) is the collection of all k-dimensional subspaces of Hθ1 (Yf ). Let ξ belong to B such that ξ 6= θ. Define F to be the linear span of {ei(ξ−θ)·x φj (θ)}kj=1 in Hξ1 (Yf ) and let u belong to F . Then k∇uk2L2 (Y

f)

‚ ‚2 = ‚∇ξ−θ u ˜‚L2 (Y

f)

+ 2Re(∇˜ u, i(ξ − θ)˜ u) + |θ − ξ|2 k˜ uk2L2 (Y ) f ´ ≤ µ2k (θ) + 2µk (θ)|θ − ξ| + |ξ − θ|2 × kuk2L2 (Y ) , =

k∇˜ uk2L2 (Y ) f

`

f

where ∇θ = ∇ + iθ, and u ˜ is a linear combination of {φj (θ)}kj=1 . Since F has dimension k in Hξ1 (Yf ) we conclude from (19) µ2k (ξ) ≤ µ2k (θ) + 2µk (θ)|θ − ξ| + |ξ − θ|2 . (20) Switching the roles of θ and ξ allows us to conclude (i). We now prove (ii), beginning with the proof of k→∞

0 ≤ inf µ21 (θ) ≤ inf µ22 (θ) ≤ . . . inf µ2k (θ) ≤ . . . → ∞. θ∈B

θ∈B

θ∈B

(21)

Observe that {inf µ2k (θ)}∞ k=1 is a non-decreasing sequence. To see that its only accumulations point is at θ

∞, we let θk be the value for which µ2k (θ) achieves its minimum. Then from (20) we conclude µk (θ) ≤ µk (θk ) + C = inf µk (θ) + C. θ

Here C is a constant independent of k. Passing to the limit in k, we conclude (21).

14

Now consider the max-min characterization of the principle eigenvalue µ21 (θ) =

k∇uk2L2 (Y

f)

min u∈H 1 (Yf ) θ

u6=0

kuk2L2 (Y

.

(22)

f)

Taking test function eiθ·x belonging to Hθ1 (Yf ) in (22) we conclude µ21 (θ) ≤ |θ|2 , hence inf µ21 (θ) = 0 = µ1 (0).

θ∈B

We note from system (6) that µ22 (θ) > 0

for θ ∈ (0, 2π)d

for if µ22 (θ) is zero at some θ ∈ (0, 2π)d , then its corresponding eigenvector must be zero due to the θ-quasi periodic boundary conditions. Thus it remains to show µ22 (θ) is positive for θ = 0. We have the variational formula µ22 (0)

=

min

max

F ∈Σ2 (0) u∈F

k∇uk2L2 (Y

f)

kuk2L2 (Y

f)

=

max

min

F ∈Σ1 (0) u∈F ⊥

k∇uk2L2 (Y

f)

kuk2L2 (Y

≥

f)

k∇uk2L2 min

1 (Y )/R u∈H# f

#

kuk2L2

#

(Yf )

> 0,

(Yf )

where the inequality follows by taking the subspace F to be the constants. The proof of (ii) is now complete. For the proof of (iii), notice that (20) implies the inequalities µ1 (θ) ≤ |θ|,

(23)

µ2 (0) ≤ µ2 (θ) + |θ|. If we take the ball BR1 (0) = {θ ∈ Rd : |θ| < R1 },

R1 = µ2 (0)/2,

then the inequalities in (23) imply µ1 (θ) < µ2 (θ)

on BR1 (0).

Hence, our claim about the simplicity of the principle eigenvalue follows. To prove (iv) and (v), we start by explaining the construction of the φk . We recall the Poincare inequality. For θ ∈ (0, 2π)d there exists a constant C(θ) > 0 such that kuk2 ≤ C(θ) k∇uk2

∀u ∈ Hθ1 (Yf )

(24)

Let’s assume for the moment that (24) holds. Then given f ∈ L2θ (Yf ), there exists a unique u ∈ Hθ1 (Yf ) satisfying 8 −4u = f in Yf > > > < ∂u (25) =0 on ∂Ys > ∂ν > > : iθ·p d u(x + p, θ) = e u(x, θ), ∀p ∈ Z i.e.

Z

∇u(x, θ) · ∇σ(x, θ)∗ dx = Yf

Z

f (x)σ(x, θ)∗ dx

Yf

15

` ´ ∀σ ∈ Hθ1 (Yf ), k∇·k2 .

Now, we define the mapping K

θ L2θ (Yf ) 3 f −→ u ∈ L2θ (Yf ).

The mapping Kθ is linear and continuous. Using (25) we can easily deduce the self-adjointness of Kθ . Kθ is also a compact operator as a consequence of the Poincare inequality (24) and Sobolev compactness. It follows from Hilbert-Schmidt theorem that there exist φk (θ), µ2k (θ), k = 1, 2, 3, . . . satisfying (6) such that the functions φk (θ) form an orthonormal basis of L2θ (Yf ). Using the weak formulation of the Neumann n o ` ´ ∇φk (θ)}k ∞ problem (6) we conclude forms an orthonormal basis of Hθ1 (Yf ), k∇·k2 . µ (θ) k

k=1

When θ = 0, we have the classical periodic case. We note that the principle eigenvalue is zero, and its corresponding eigenvector is a constant. Then to obtain the remaining eigenvectors and eigenvalues of (6), we look at the space L2# (Yf )/R and we argue as above. We now prove (24). For this we appeal to the closed graph theorem. We first remark ` 1 ´ Hθ (Yf ), k∇.k2

(26)

is a Banach space for each θ ∈ (0, 2π)d . Indeed, what is required to show is the completeness, since it is a normed space. Let um be a Cauchy sequence in (26). Then there exists v ∈ L2θ (Yf ) such that ∇um → v in L2 (Yf ). To see v is a gradient we appeal to the following well known lemma, a proof of which can be found in [26] (see Remark 1.9). Lemma 4.1 Let θ ∈ [0, 2π)d and f (θ) belong to L2θ (Yfε )d . Suppose that Z

f (θ, x) · σ(θ, x)∗ dx = 0 Yfε

for all σ(θ) ∈ L2θ (Yf )d satisfying

(

divσ = 0

in Yfε

σ·ν =0

on ∂Ysε .

(27)

Then there exists g(θ) ∈ Hθ1 (Yfε ) such that f (θ) = ∇g(θ). Now take σ ∈ L2θ (Yf )d , divergence free in x and θ-quasi periodic. Then Z Yf

v · σ ∗ = lim

m→∞

Z

∇um · σ ∗ = 0. Yf

It follows that v = ∇p for some p ∈ Hθ1 (Yf ). Since the space defined in (26) is a Banach space, one can apply the closed graph theorem to the identity mapping i:

„ “ ”1/2 « ` ´ Hθ1 (Yf ), k·k22 + k∇.k22 → Hθ1 (Yf ), k∇.k2

to conclude that i−1 is continuous, hence (24) follows and the proof of Proposition 2.5 is now complete. 

16

4.2

Proof of Proposition 2.8:

M We begin with the proof of the first inclusion in (10) and inequality (11). Assume f belongs to Skε ; we need to show that [f ]2 belongs to H(Rdε ). Notice that 1 > > χ0 · ν = 0 on ∂Ys > > > : y → χ0 is Y -periodic

(36)

We now study the limit of f2ε , and denote its two-scale limit by χ1 . Making use of the definition of the projections mappings Πε±1 , we obtain Z f2ε (x) = 2 − B ε (θ) [ Φε1 (x, θ) ]2 ,

(37)

B

where ε B ε (θ) = (f# (θ), [Φε1 (θ)]2 )L2 (Y ε ) . f

Taking the divergence of (37) yields Z 2 − µ1 (θ) B ε (θ) [ Φε1 (x, θ)]1 i B Z X 2 = − µ c1 (p)eiθ·p B ε (θ) [ Φε1 (x, θ) ]1 i B d

ε divf2ε (x) =

p∈Z

=

2 X µ c1 (p) hε (x + εp), i d p∈Z

20

(38)

ˆ ˜ where the last equality in (38) made use of the quasi-periodicity of Φε1 1 . Here hε is defined as Z hε (x) = − B ε (θ) [ Φε1 (x, θ) ]1 . B

We now pass to the limit in (38), and we begin with the left-hand side. Note that f2ε has a normal ∞ (Y )] we have component which vanishes on ∂Rdε , hence for σ ∈ D[Rd ; C# Z

“ x” divf2ε σ x, ε Rd Z ε h “ x” “ x ”i f2ε · ε ∇x σ x, = + ∇y σ x, ε ε Rd Zε ε→0 → χ1 · ∇y σ.

−ε

Rd ×Yf

In order to pass to the limit in the right hand side of (38), we first notice that if we denote the two-scale limit of hε by h, we have 2−s

hε (x + εp) * h,

∀ p ∈ Zd .

(39)

Indeed, from a change of variables we have the equality Z Rd ε

Z “ “ x” x” hε (x + εp) σ x, dx = hε (x) σ x − εp, dx, ε ε Rd ε

and passing to the limit in ε gives us (39). It follows that Z “ x” 2 X µ c1 (p) hε (x + εp)σ x, d i ε Rε p∈Zd Z X ε→0 2 hσ → µ c1 (p) i Rd ×Yf p∈Zd Z 2 h σ. = µ1 (0) i Rd ×Yf Since µ1 (0) = 0, we deduce the following two-scale system for χ1 8 > < divy χ1 = 0 in Yf χ1 · ν = 0 on ∂Ys > : y → χ1 is Y -periodic

(40)

To compute the two-scale limit of QεM f ε , which we denote by χ2 , we recall from proposition 2.8 that QεM f ε = ∇g ε and satisfies the inequality kg ε k2 ≤ Cε k∇g ε k2 .

(41)

∞ (Y )]d satisfying σ = 0 on Y and div σ = 0 we have For any σ ∈ D[Rd ; C# s y

Z − Rd ε

“ x” “ x” Z = g ε divx σ x, . QεM f ε · σ x, ε ε Rd ε

21

(42)

Using (41), we pass to the limit in (42) and deduce Z Rd ×Yf

χ2 · σ(x, y) = 0.

The functions that are orthogonal to divergence free functions are exactly gradients. We deduce that there exist a unique function g0 ∈ L2 (Rd ; H# (Yf )/R) such that χ2 = ∇g0 . If we denote the two-scale limit of f ε by f0 , then combining all the two-scale limits above we have f 0 = χ0 + χ1 + χ2 . Since the decomposition f0 = P y f0 + Q y f0 is unique the proof of theorem 3.4 is complete.  Remark 4.6 As noted in the proof of theorem 3.4, we have the equality Py a0 = χ0 + χ1 where χ0 and χ1 satisfy (36), (40) respectively. We can say slightly more about χ1 . Indeed, notice that the curl hence

ˆ`

´ ˜ Πε1 + Πε−1 (0, aε ) 2 = 0, Z curlx − χ1 = 0 Yf

curly χ1 = 0 By elliptic regularity, it follows that χ1 is regular in y. In fact, more can be said about χ1 in dimension d = 2. We define V = {u ∈ L2# (Yf )d ; div u = 0 on Yf , u · ν = 0 on ∂Ys } and Γ = curl. Then the kernel of Γ is a Banach space of dimension N + 1 where N equals the number of holes in Yf . We refer the reader to [17] for the details.

5

Proof of the convergence Results

The existence of a solution (pε , uε ) ∈ C([0, T ); L2 (Rdε ) × L2 (Rdε )d ) to the linear system (1) is very classical and we do not recall it here. We only observe that the solution (pε , uε ) satisfies kpε (t)k2L2 (Rd ) + kuε (t)k2L2 (Rd ) + 2εβ ε

ε

t

Z 0

k∇uε k2L2 (Rd ) = kbε k2L2 (Rd ) + kaε k2L2 (Rd ) . ε

ε

ε

We note that the proofs of the main results repeatedly make use of the Poincar´ e inequality and the energy estimate satisfied by the sequence of solutions (pε , uε ). We state the Poincar´ e inequality on the domain Rdε (see [24]) Theorem 5.1 Let f ∈ H01 (Rdε ). Then we have kf kL2 (Rd ) ≤ Cε k∇f kL2 (Rd ) . ε

ε

22

5.1

(1 < β < 2) Proof of theorem 3.6: We introduce the following error terms 0

1 « x A − qε = p − b(x) − @ ε pj , x, ε2−β ε j=0 0 1 « „ l X t x ε ε 2j(β−1) A − vε βl = u − @ ε uj , x, 2−β ε ε j=0

αεl

ε

l X

„

(2j+1)(β−1)

t

(43)

where (p0 , u0 ) = (p, u) is the unique solution to (16), and (pj , uj ), j ≥ 1, satisfy the coupled two-scale system 8 ∂τ uj − 4y uj + ∇y pj = 0 in Rd × Yf > > > > > divy uj = −∂τ pj−1 > > > > > uj = 0 on ∂Ys > < Z (44) pj = 0 > > > Y > f > > > > > uj (τ = 0) = 0 > > : y → (uj , pj ) is Y -periodic We present an analysis of the error term (43) for l = 0 which is enough if β ∈ (4/3, 2). At the end of the subsection, we will conclude with a discussion on the adjustments needed to justify the error terms in the case 0 < l for β ∈ (1, 4/3]. For l = 0 the proposed error term (αε0 , β0ε ) = (αε , β ε ) satisfies the following initial/boundary value problem 8 ∂t αε + divβ ε = F ε (t, x) > > > > > ∂ β ε − εβ 4β ε + ∇αε = Gε (t, x) > > > t < βε = 0 > > > αε (t = 0) = bε1 − b(x) − εβ−1 p0 (τ = 0) > > > “ x” > > : ε ε β ε (t = 0) = PM a (x) − Py a0 x, ε

in (0, ∞) × Rdε on ∂Rdε

(45)

where the right-hand side terms are F ε (t, x) = −ε2β−3 ∂τ p − divx u, Gε (t, x) = εβ 4x u + 2 εβ−1 divy ∇x u − ∇b + εβ−1 ∇x p. Remark 5.2 Due to the parabolic regularization of the Stokes equation (16) in the variable y, the trace “ ” t x , x, are well defined for k = 0, 1 and |α|, |β| ≤ 2. ε ε2−β

of the terms ∂τk ∂xα ∂yβ (p, u)

The energy equality associated to system (45) reads kαε (t)k2L2 (Rd ) + kβ ε (t)k2L2 (Rd ) + 2εβ ε

Z

ε

kαε (0)k2L2 (Rd ) + kβ ε (0)k2L2 (Rd ) + 2 ε

ε

and we now estimate the term

t

Z 0

Z

t

0 tZ

Z 0

k∇β ε (s)k2L2 (Rd ) = ε

(46) (F ε αε + Gε · β ε ) , Rd ε

(F ε αε + Gε · β ε ) . Rd ε

23

We write

Z Rd ε

F ε αε = F1 + F2 ,

and we obtain the following bounds t

Z t x ”‚ ‚2 , x, ds + kαε k2L2 (Rd ) ‚ ε ε2−β ε L2 (Rdε ) 0 0 Z ∞‚ Z t “ ”‚ x ‚2 ‚ ≤ ε3β−4 dτ + kαε k2L2 (Rd ) ‚∂τ p τ, x, ‚ ε ε L2 (Rdε ) 0 0 Z ∞ Z t k∂τ p (τ, x, y)k2L2 (Rd ×Y ) dτ + kαε k2L2 (Rd ) ≤ ε3β−4 Z

|F1 | ≤ ε4β−6

‚ “ ‚ ‚∂τ p

s

f

0

3β−4

≤ Cε

t

Z + 0

t

kαε k2L2 (Rd ) ε

(47)

ε

0

,

Z t x ”‚ ‚2 , x, ds + kαε (s)k2L2 (Rd ) ‚ ε ε2−β ε L2 (Rdε ) 0 0 Z t Z ∞‚ ”‚2 “ x ‚ ‚ dτ + kαε (s)k2L2 (Rd ) ≤ Cε2−β ‚ ‚divx u τ, x, ε ε L2 (Rdε ) 0 0 Z ∞ Z t ≤ Cε2−β kdivx u (τ, x, y)k2L2 (Rd ×Y ) dτ + kαε (s)k2L2 (Rd ) Z

|F2 | ≤ C

s

‚ “ ‚ ‚divx u

f

0

≤ Cε2−β +

t

Z

(48)

ε

0

kαε (s)k2L2 (Rd ) . ε

0

We now explain the estimates above. The first inequality in the estimates of F1 , F2 is just an application of the Cauchy-Schwartz inequality. The second inequality is the change of variable s = ε2−β τ . The third inequality in the estimates follows from strong two-scale convergence and hence holds for ε small enough. We now justify the energy bounds on ∂τ p(τ, x, y) and divx u(τ, x, y), beginning with divx u. First, notice that ∂xα u satisfies the same system (44) as u with the initial data ∂xα u(t = 0) = ∂xα Py a0 . Furthermore, we have the following energy equality k∂xα u(τ )k2L2 (Rd ×Y

f)

Z +2 0

τ

‚ α ‚ ‚∂x ∇y u(τ 0 )‚2 2 d L (R ×Y

f)

= k∂xα Py a0 k2L2 (Rd ×Y

f)

.

(49)

By the Poincar´ e inequality on the cell Yf , we also deduce that τ

Z

‚ α ‚ ‚∂x u(τ 0 )‚2 2

L (Rd ×Yf )

0

Hence,

∞

Z 0

Z ≤C 0

τ

‚ α ‚ ‚∂x ∇y u(τ 0 )‚2 2 d L (R ×Y

f)

kdivx u (τ, x, y)k2L2 (Rd ×Y

f)

.

(50)

≤ C.

To obtain the energy bound for ∂τ p(t, x, y), we begin by taking a time derivative of the vector equation in (44) ∂τ τ u − 4y ∂τ u + ∇y ∂τ p = 0. (51) We impose the following compatibility conditions on the initial data ∂τ u(0): ∂τ u(0) = 0 = ∇y ∂τ u(0) on ∂Ys .

24

Multiplying (51) by ∂τ u and ∂τ τ u and integrating by parts over Yf yields the equalities τ

Z

k∂τ u(τ )k2L2 (Rd ×Y ) + 2 f Z τ ‚ ‚2 0 ‚∂τ 0 τ 0 u(τ )‚ 2 2

L (R×Yf

0

‚ ‚ ‚∇∂τ 0 u(τ 0 )‚2 2

L (Rd ×Yf )

0

= k∂τ u(0)k2L2 (Rd ×Y

f)

, (52)

+ k∇y ∂τ u(τ )k2L2 (R×Y )

f)

= k∇y ∂τ u(0)k2L2 (R×Y

f

).

Now using equation (51), along with the energy bounds (49), we obtain the following estimate τ

Z

≤C

L (Rd ×Yf )

0

τ

Z

‚ ‚ ‚∂τ p(τ 0 )‚2 2

‚ ‚ ‚∇y ∂τ 0 p(τ 0 )‚2 2

−1 L (Rd ;H# (Yf ))

0

≤ C k∂τ u(0)k2L2 (Rd ) . + k∇y ∂τ u(0)k2L2 (R×Y

f)

ε

.

Similarly, we write t

Z

Z Rd ε

0

Gε · β ε = G1 + G2 + G3 + G4

and we have the following estimates «1/2 „Z t «1/2 x ”‚ ‚2 kβ ε (s)k2L2 (Rd ) ‚ ε ε ε L2 (Rdε ) 0 0 « „Z „Z t ‚ «1/2 1/2 “ s t x ”‚ ‚2 ‚ ≤ Cεβ+1 k∇β ε (s)k2L2 (Rd ) ‚ ‚4x u 2−β , x, ε ε ε L2 (Rdε ) 0 0 Z t Z t‚ ”‚ “ s β 2 x ‚ ε ‚ + k∇β ε (s)k2L2 (Rd ) ≤ Cε2+β ‚ ‚4x u 2−β , x, ε ε ε L2 (Rdε ) 5 0 0 Z t β ε ≤ Cε4 + k∇β ε (s)k2L2 (Rd ) , ε 5 0

|G1 | ≤ Cεβ

„Z

t

s

‚ “ ‚ ‚4x u

, x, 2−β

«1/2 „Z t «1/2 x ”‚ ‚2 kβ ε (s)kL2 (Rd ) ‚ ε ε ε L2 (Rdε ) 0 0 „Z t ‚ «1/2 „Z t «1/2 “ s ”‚ x ‚2 ‚ ≤ Cεβ k∇β ε (s)kL2 (Rd ) ‚divy ∇x u 2−β , x, ‚ ε ε ε L2 (Rdε ) 0 0 Z t‚ ”‚2 “ s β Z t x ε ‚ ‚ + k∇β ε (s)k2L2 (Rd ) ≤ Cεβ ‚ ‚divy ∇x u 2−β , x, ε ε ε L2 (Rdε ) 5 0 0 Z t β ε ≤ Cε2 + k∇β ε (s)k2L2 (Rd ) , ε 5 0 „Z t «1/2 „Z t «1/2 kβ ε (s)k2L2 (Rd ) |G3 | ≤ C k∇bk2L2 (Rd )

|G2 | ≤ Cεβ−1

„Z

t

‚ “ ‚ ‚divy ∇x u

s

, x, 2−β

ε

0

t

„Z

k∇bk2L2 (Rd )

≤ Cε

ε

0

≤ Cε2−β

t 0

t

k∇β ε (s)k2L2 (Rd )

Z εβ t k∇bk2L2 (Rd ) + k∇β ε (s)k22 ε 5 0 0 Z εβ t + k∇β ε (s)k22 , 5 0

Z

≤ Ctε2−β

ε

0

«1/2 „Z

25

ε

«1/2

«1/2 «1/2 „Z t x ”‚ ‚2 kβ ε (s)k2L2 (Rd ) ‚ ε ε L2 (Rdε ) 0 0 «1/2 „Z t «1/2 „Z t ‚ “ s ”‚ 2 x ‚ ‚ k∇β ε (s)k2L2 (Rd ) ≤ Cεβ ‚∇x p 2−β , x, ‚ ε ε ε L2 (Rdε ) 0 0 Z t Z ∞‚ “ ”‚ β x ‚2 ε ‚ ≤ Cε2 + k∇β ε (s)k2L2 (Rd ) ‚∇x p τ, x, ‚ ε ε L2 (Rdε ) 5 0 0 β Z t ε k∇β ε (s)k2L2 (Rd ) . ≤ Cε2 + ε 5 0

|G4 | ≤ Cεβ−1

„Z

t

‚ “ ‚ ‚∇x p

s

ε2−β

, x,

Here, the second inequality in the estimates of G1 , G2 made use of the Poincar´ e inequality on the domain Rdε . We concluded the final inequality in the estimates G1 , G2 with a change of variable in time along with the bounds in (49), (50). Combining all the estimates on the force terms, the energy bound (46) becomes kαε (t)k2L2 (Rd ) + kβ ε (t)k2L2 (Rd ) + ε

≤ kα where

ε

ε

(0)k2L2 (Rd ) ε

+ kβ

ε

εβ 5

t

Z 0

k∇β ε (s)k2L2 (Rd )

(0)k2L2 (Rd ) ε

ε

Z

t

+ 0

(53)

kα (s)k2L2 (Rd ) + Ct,ε , ε

ε

“ ” Ct,ε = C ε3β−4 + ε2−β + ε4 + ε2 + tε2−β .

Hence, by applying Gronwall’s lemma to (53), we conclude the proof of Theorem 3.6 for β ∈ (4/3, 2) To extend this result to β ∈ (1, 4/3], we take l > 0 large enough such that (4l + 3)(β − 1) − 1 > 0 and proceed with a similar analysis as above. Notice that we have ε(4l+4)(β−1)−2

t

Z 0

Z Rd ε

“ ” |∂τ pl |2 = O εml (β−1)−1

where ml = 4l + 3, hence the proof of theorem 3.6 is now complete. 

5.2

(β = 2) Proof of theorem 3.8: We define the following error terms “ “ x” x” − εp2 (t, x) − ε2 p3 t, x, − qε , αε = pε − p(t, x) − εp1 t, x, ε ε “ “ x” x” β ε = uε − u t, x, − εu1 t, x, − vε , ε ε

where (p2 (t, x), p3 (t, x, y), u1 (t, x, y)) uniquely satisfies the two-scale system Z 8 > > |Yf |∂t p2 + divx u1 = 0 in Rd > > > Yf > > > > > ∂t u1 − 4y u1 + ∇x p2 + ∇y p3 = −∇x p1 in Rd × Yf > > > Z > > > > > divy u1 = divx u − divx u > > > Yf > < u1 = 0 on ∂Ys > Z > > > > p3 = 0 > > > > Yf > > > > > p(t = 0) = 0 > > > > > u(t = 0) = 0 > > > : y → (p3 , u1 ) is Y -periodic .

26

(54)

(55)

The system associated to the error term (αε , β ε ) reads 8 ∂t αε + divβ ε = F ε (t, , x) > > > > ε β > > ∂t β − ε 4β ε + ∇αε = Gε (t, x) > > < βε = 0 > > ε > α (t = 0) = bε1 − b(x) − εp1 (t = 0) − ε2 p3 (t = 0) > > > “ x” > > ε ε : β ε (t = 0) = PM a (x) − Py a0 x, ε

in (0, ∞) × Rdε on ∂Rdε

where the force terms here are given by F ε (t, x) = −ε∂t p1 − ε2 ∂t p3 Gε (t, x) = ε2 4x u + 2ε divy ∇x u + ε3 4x u1 + 2ε2 divy ∇x u1 − ε2 ∇x p3 . We now imitate the proof given for the case β ∈ (1, 2). Notice that (αε , β ε ) satisfies the energy bound in (46) (for β = 2), hence it suffices to show ˛Z Z ˛ ˛ t ˛ ˛ ˛ F ε αε + G ε · β ε ˛ ˛ ˛ 0 Rd ˛ ε (56) Z t Z t 2 2 γ 2 ε ε ≤ O (ε ) + C1 ε k∇β (s)kL2 (Rd ) + C kα (s)kL2 (Rd ) , ε

0

ε

0

for some 0 < γ and constant C1 < 1. We begin by writing Z Rd ε

F ε αε = F1 + F2 ,

and we obtain the following bounds Z t Z ∞‚ “ x ”‚ ‚2 ‚ + kαε k2L2 (Rd ) |F1 | ≤ ε2 ‚ ‚∂τ p1 t, x, ε ε L2 (Rdε ) 0 0 Z ∞ Z t ≤ ε2 k∂τ p1 (t, x, y)k2L2 (Rd ×Y ) + kαε k2L2 (Rd ) f

0

≤ Cε2 +

t

Z

0

ε

kαε k2L2 (Rd ) , ε

0

(57)

∞

Z t ‚ “ x ”‚ ‚2 ‚ + kαε k2L2 (Rd ) |F2 | ≤ ε4 ‚ ‚∂τ p3 t, x, ε ε L2 (Rdε ) 0 0 Z ∞ Z t ≤ ε4 k∂τ p3 (t, x, y)k2L2 (Rd ×Y ) + kαε k2L2 (Rd ) Z

f

0

≤ Cε4 +

t

Z 0

0

ε

kαε k2L2 (Rd ) . ε

We now justify the energy bound on ∂t p1 (τ, x, y) (the energy bound for ∂t p3 is obtained similarly). −1 Notice, it suffices to show that ∇y ∂t p1 is bounded in L2 ((0, ∞) × Rd ; H# (Yf ))d . We begin by taking a time derivative of the vector equation in (17) ∂tt u − 4y ∂t u + ∇x ∂t p + ∇y ∂t p1 = 0.

(58)

Multiplying (58) by ∂t u and ∂tt u and integrating by parts over Yf yields the equality k∂t u(t)k2L2 (Rd ×Y

f

=

+ |Yf | k∂t p(t)k2L2 (Rd ) + 2 )

k∂t u(0)k2L2 (Rd ×Y ) f

+

|Yf | k∂t p(0)k2L2 (Rd )

27

,

t

Z 0

k∇y ∂s u(s)k2L2 (Rd ×Y

f)

(59)

and the inequality t

Z 0

= ≤

k∂ss u(s)k2L2 (Rd ×Y

f)

+

1 k∇y ∂t u(0)k2L2 (Rd ×Y f 2

1 k∇y ∂t u(t)k2L2 (Rd ×Y ) f 2 Z tZ ∇x ∂t p · ∂tt u − )

(60)

Rd ×Yf

0

|Yf | 1 k∇y ∂t u(0)k2L2 (Rd ×Y ) + f 2 2

t

Z 0

k∇x ∂s p(s)k2L2 (Rd ×Y

f)

+

1 2

Z

t

0

k∂ss u(s)k2L2 (Rd ×Y

f)

.

Next, we observe that ∂xα (p, p1 , u) satisfies the same system (17) as (p, p1 , u) with the initial data ∂xα (b, Py a0 ). Moreover, we have the following energy equality k∂xα u(t)k2L2 (Rd ×Y

f)

=

+ |Yf | k∂xα p(t)k2L2 (Rd ) + 2

|Yf | k∂xα bk2L2 (Rd ×Y ) f

+

k∂xα Py a0 k2L2 (Rd )

Z

t

0

k∇y ∂xα u(s)k2L2 (Rd ×Y

f)

(61)

,

Taking a time derivative of the scalar equation in (17) yields |Yf |∂xα ∂t p + ∂xα divx

Z u = 0, Yf

and we deduce the inequality t

Z 0

k∂xα ∂s p(s)k2L2 (Rd ) ≤ C

Z

t

Z ≤C 0

k∂xα divx u(t)k2L2 (Rd ×Y

f)

0 t

(62) k∂xα divx ∇y u(t)k2L2 (Rd ×Y

f

. )

Here, the last inequality in (62) follows from the Poincar´ e inequality in the cell Yf . Combing the estimates (59), (60), (61), (62) we conclude k∇y ∂t p1 kL2 ((0,∞)×Rd ;H −1 (Y #

f ))

≤C

Similarly, we write t

Z 0

Z Rd ε

Gε · β ε = G1 + G2 + G3 + G4 + G5

and we have the following estimates

28

«1/2 „Z t «1/2 ‚ “ x ”‚ ‚2 ‚ kβ ε (s)k2L2 (Rd ) ‚ ‚4x u t, x, ε ε L2 (Rdε ) 0 0 „Z ∞ ‚ «1/2 «1/2 „Z t “ ”‚ x ‚2 ‚ ≤ Cε3 k∇β ε (s)k2L2 (Rd ) ‚4x u t, x, ‚ ε ε L2 (Rdε ) 0 0 Z ∞‚ “ ”‚2 2 Z t x ε ‚ ‚ k∇β ε (s)k2L2 (Rd ) ≤ Cε4 + ‚4x u t, x, ‚ ε ε L2 (Rdε ) 6 0 0 2 Z t ε ≤ Cε4 + k∇β ε (s)k2L2 (Rd ) , ε 6 0 „Z ∞ ‚ «1/2 „Z t «1/2 “ x ”‚ ‚ ‚2 |G2 | ≤ Cε kβ ε (s)kL2 (Rd ) ‚ ‚divy ∇x u t, x, ε ε L2 (Rdε ) 0 0 «1/2 „Z ∞ ‚ «1/2 „Z t ”‚ “ x ‚2 ‚ k∇β ε (s)kL2 (Rd ) ≤ Cε2 ‚divy ∇x u t, x, ‚ ε ε L2 (Rdε ) 0 0 Z ∞‚ “ ”‚2 2 Z t x ε ‚ ‚ ≤ Cε2 + k∇β ε (s)k2L2 (Rd ) ‚ ‚divy ∇x u t, x, ε ε L2 (Rdε ) 6 0 0 Z t 2 ε ≤ Cε2 + k∇β ε (s)k2L2 (Rd ) , ε 6 0 „Z ∞ ‚ «1/2 „Z t «1/2 ”‚ “ x ‚2 ‚ |G3 | ≤ Cε3 kβ ε (s)k2L2 (Rd ) ‚ ‚4x u1 t, x, ε ε L2 (Rdε ) 0 0 «1/2 „Z t «1/2 „Z ∞ ‚ ”‚ “ 2 x ‚ ‚ k∇β ε (s)k2L2 (Rd ) ≤ Cε4 ‚ ‚4x u1 t, x, ε ε L2 (Rdε ) 0 0 Z ∞‚ Z t ”‚ “ 2 2 ε x ‚ ‚ ≤ Cε6 + k∇β ε (s)k2L2 (Rd ) ‚ ‚4x u1 t, x, ε ε L2 (Rdε ) 6 0 0 Z t 2 ε ≤ Cε6 + k∇β ε (s)k2L2 (Rd ) , ε 6 0 „Z ∞ ‚ «1/2 „Z t «1/2 “ x ”‚ ‚2 ‚ |G4 | ≤ Cε2 kβ ε (s)kL2 (Rd ) ‚ ‚divy ∇x u1 t, x, ε ε L2 (Rdε ) 0 0 «1/2 „Z t «1/2 „Z ∞ ‚ ”‚ “ x ‚2 ‚ k∇β ε (s)kL2 (Rd ) ≤ Cε3 ‚ ‚divy ∇x u1 t, x, ε ε L2 (Rdε ) 0 0 Z ∞‚ ”‚2 “ 2 Z t ε x ‚ ‚ ≤ Cε4 + k∇β ε (s)k2L2 (Rd ) , ‚ ‚divy ∇x u1 t, x, ε ε L2 (Rdε ) 6 0 0 Z t 2 ε ≤ Cε4 + k∇β ε (s)k2L2 (Rd ) , ε 6 0 „Z t ‚ «1/2 „Z t «1/2 “ ”‚ x ‚2 ‚ |G5 | ≤ Cε2 kβ ε (s)k2L2 (Rd ) ‚∇x p3 t, x, ‚ ε ε L2 (Rdε ) 0 0 «1/2 „Z t «1/2 „Z t ‚ “ ”‚ 2 x ‚ ‚ ≤ Cε3 k∇β ε (s)k2L2 (Rd ) ‚∇x p3 t, x, ‚ ε ε L2 (Rdε ) 0 0 Z ∞‚ Z t “ ”‚ 2 2 x ‚ ε ‚ ≤ Cε4 k∇β ε (s)k2L2 (Rd ) + ‚∇x p3 t, x, ‚ ε ε L2 (Rdε ) 6 0 0 Z t 2 ε k∇β ε (s)k2L2 (Rd ) . ≤ Cε4 + ε 6 0 |G1 | ≤ Cε2

„Z

∞

We remark that the final bounds in G1 , G2 make use of (61) and the Poincar´ e inequality on the cell Yf . We do not provide the energy bounds for the terms 4x u1 , divy ∇x u1 , ∇x p3 , but note that they can be obtained similarly to the methods presented here.

29

Combing all the estimates associated with the scalar and vector force terms, (56) holds for γ = 2 and C1 = 5/6. The proof of Theorem 3.8 is now complete. 

5.3

(2 < β) Proof of theorem 3.10:

In the case 2 < β, there is a boundary layer in space of size εβ/2 associated to the micro-incompressible part of the flow. The asymptotic analysis necessary in the case 2 < β is interesting in its own right. We will only detail the construction of the complete asymptotic analysis in the case 2 < β. The convergence proof based on the estimate of the error terms (αε , β ε ) follows exactly as in the previous two proofs and will be omitted. We introduce the following asymptotic expansions „ « “ ”« “x” “ d( xε ) x x” + pbdy t, x, Π , (β−2)/2 χ pk,l t, x, k,l ε ε ε ε k,l „ « « „ x “ “ “ ” ” X d( ε ) x x” x t, x, Π χ uε − v ε ∼ εk+l(β−2)/2 uk,l t, x, + ubdy , (β−2)/2 k,l ε ε ε ε k,l pε − q ε ∼

X

εk+l(β−2)/2

„

(63)

Here, Πy is the projection of y on the boundary ∂Ys and d(y) denotes the distance function to the boundary, namely the distance between y and ∂Ys . Note that Π is well defined in a small neighborhood of the boundary ∂Ys . The function χ(y) ∈ C0∞ (Y ) is a smooth cut-off function such that χ(y) = 1 in a neighborhood of Ys and χ(y) = 0 if d(y) > δ for some δ small enough. Here δ > 0 is taken in such a way that Π is uniquely determined in the neighborhood {y | 0 < d(y) < δ} and ∇d is a unit normal vector on supp(χ) ∩ Yf . We will use ν = ∇d“to denote the ” inward normal vector to Yf . An important remark to bdy pbdy when y k,l , uk,l x χ( ε ) in (63). “ ” bdy pbdy (t, x, Πy, ξ) k,l , uk,l

keep in mind is that we only need presence of the cut-off function The boundary layer terms

“

bdy pbdy k,l , uk,l

is close to the boundary ∂Ys which justifies the

satisfy

”

→

0

(64)

when ξ goes to ∞ and uk,l (t, x, y) + ubdy k,l (t, x, Πy, ξ = 0) = 0

when y = Πy ∈ ∂Ys .

(65)

We will also need a curvilinear coordinate system Ξ = (ξ 0 , ξd ), ξ 0 = (ξ1 , ..., ξd−1 ) adapted to ∂Ys and defined in a neighborhood of ∂Ys such that ∂Ys = {ξd = 0} and Yf is located at the side {ξd > 0}. Of course, we think here of Ξ → y(Ξ) as a change of variables where the expansion of the operators ∆, ∇ and div will have simpler expressions. We will follow very similar notations to those in [14, 13]. We define the tubular neighborhood Ys,δ = {−δ < ξd < δ}. We also assume that our curvilinear coordinate system (ξ 0 , ξd ) is orthogonal and that « d „ X ∂y ∂y ∂yi 2 · = =1 ξd ξd ξd 1 which means that ξd = d(y) = ε(β−2)/2 ξ in Ys,δ ∩ Yf . We introduce the vectors gα =

∂y = ξα

„

∂y1 ∂yd , ..., ξα ξα

« ,

1 ≤ α ≤ d.

Hence, the metric (gαβ )1≤α,β≤d = (gα · gβ )1≤α,β≤d = diag(g11 , ..., gdd ) and gdd = 1. We will use the p √ notation hi = gii > 0, i = 1, ..., d − 1 and h = det((gαβ )1≤α,β≤d ). With these notations, we can also

30

rewrite divy U , ∆y f and ∇y f in the following way. If U and f are defined in a neighborhood of ∂Ys and P such that U = di=0 Ui ei where ei = |ggi | and f is scalar valued, then i

divy U =

∆y f =

1 h

„ « d−1 h 1 X ∂ 1 ∂(hUd ) , Ui + h i=1 ∂ξi hi h ∂ξd

X 1≤i,j≤d−1

∇y f =

∂ ∂ξi

d−1 X i=1

„

h ∂f h2i ∂ξi

« +

(66)

h0 ∂f ∂2f + , h ∂ξd ∂ξd2

(67)

1 ∂f ∂f ed . ei + hi ∂ξi ∂ξd

(68)

P If U is a vector given by U = di Ui ei , then computing ∆y U is more complicated. We will not give the exact formula here. We refer to formula (2.17) in [13]. The first term on the right-hand side of (66) will be denoted divtg y (U ). For a vector field which is in the tangent space to ∂Ys , it corresponds to the divergence on the manifold ∂Ys . We will also use the notation U tg to denote the tangential part of the verctor field U , namely U tg =

d−1 X

Ui ei .

(69)

i=1

If we consider now functions that depend on x and y, namely of the form f (x, we also have the following expansions of ∇, div and ∆, namely ∇f = ∇x f + divu = divx u + ∆f = ∆x f +

x ) ε

and u(x,

x ), ε

then

1 tg 1 ∇ f + β/2 (∂ξ f )∇d, ε y ε

(70)

1 1 divtg y u + β/2 ∂ξ (u.∇d), ε ε

(71)

1 tg 1 1 ∇x · ∇tg y f + 2 ∆y f + β/2 ∇d · ∇x ∂ξ f ε ε ε 1 1 h0 + β/2 ∂ξ f + β ∂ξ ∂ξ f. h ε εε

(72)

One can then write an expansion of h0 /h, namely ∞

X h0 0 h0 0 (β−2)/2 h0 εi(β−2)/2 ∂ξi d ( )(ξ 0 , 0)ξ i . (ξ , ξd ) = (ξ , ε ξ) = h h h i=0 Plugging this expansion in the definition of (72), we deduce an expansion of ∆f in powers of ε. The main observation is that it reads ∆f = ε1β ∂ξ ∂ξ f + l.o.t where l.o.t denotes terms with lower order, namely termsPwith higher powers of ε. It is not difficult to see that the same holds in the vector case, namely if U = di Ui ei then 1 1 h0 ∆U = β ∂ξ ∂ξ U i ei + β/2 (ξ 0 , 0)∂ξ U i ei + l.o.t. (73) ε h εε We included the second term on the right-hand side of (73) since it is easy to compute but it will not be crucial for the construction process. We plug this expansion into the original system (1) and then we gather the terms of the same order in the boundary layer and in the interior. When we do this gathering, we do not specify the value of β since we want to find an expansion which is valid for all values of β. By doing so, we hope to get successive systems to solve that allow us to determine the full expansion (63) (check). We will present the construction in an inductive way starting from l = 0.

31

To perform the construction of our expansion we will make the following consistency condition: bε (x) = b(x) and aε (x) = a0 (x, xε ) where a0 = Py a0 and a0 (x, y) = 0 when y ∈ ∂Ys . Moreover, we assume that b and a0 are smooth enough. Let us point out that if we only assume that a0 (x, y) · ν = 0 when y ∈ ∂Ys , then we need to incorporate an initial layer in time. This will not be done here (see [27] for more about this). Step 0: The case l = 0 bdy We deduce from the order ε−β/2 in the second equation of (1) that ∂ξ pbdy 0,0 = 0 and hence p0,0 = 0. From bdy the order ε1−β/2 , we also deduce that ∂ξ pbdy 1,0 = 0 and hence p1,0 = 0 . It is not difficult to see that bdy from the order εk−β/2 , we also deduce that ∂ξ pbdy k,0 = 0 and hence pk,0 = 0 . In a similar from the order

ε−1+(l−1)(β−2)/2 in the second equation (1), we also deduce that pbdy 0,l = 0 for all l ≥ 0. This will be used in the Step l. bdy From the order ε−β/2 in the first equation, we get that ∂ξ ubdy 0,0 · ∇d = 0 and hence, u0,0 · ∇d = 0 and then from the boundary condition, we can deduce the normal part of u0,0 , namely we deduce that u0,0 · ν = 0 on ∂Ys . From the order εk−β/2 , we also deduce that ubdy k,0 · ∇d = 0 and hence from the boundary condition that uk,0 · ν = 0 on ∂Ys . From the order ε−1 in the interior, we deduce that divy u0,0 = 0 and ∇y p0,0 = 0 and hence p0,0 is only a function of t and x, namely p0,0 = p(t, x). We also denote u = u0,0 . From the order ε0 in the interior, we deduce that ( ∂t p + divx u + divy u1,0 = 0 in Rd × Yf (74) ∂t u + ∇x p + ∇y p1,0 = 0 in Rd × Yf R R Since, p does not depend on y and that Y divy u1,0 = ∂Y u1,0 .ν = 0, we can integrate in y the first f

f

equation and deduce that (p(t, x), p1,0 (t, x, y), u(t, x, y)) is the solution to the following two-scale system Z

8 > > > > > > > > > > > > > > > > < > > > > > > > > > > > > > > > > :

|Yf |∂t p + divx

in Rd

u=0 Yf

in Rd × Yf

∂t u + ∇x p + ∇y p1,0 = 0 divy u = 0 ∇y p = 0

(75)

u·ν =0

on ∂Ys

p(t = 0) = b(x) u(t = 0) = Py a0 (x, y) y → (u, p1 ) is Y -periodic .

Solving this system is standard and we will not detail it here. Notice that this system also determines completely ∇y p1,0 (t, x, y) which will be useful for the iteration. Now, we want to solve for R (p1,0 (t, x, y), u1,0 (t, x, y)). Recall that we have divy u1,0 + divx (u − |Y1 | Y u) = 0 and that p1,0 (t, x, y) f f R can be decomposed as p1,0 (t, x, y) = p1,0 (t, x) + p˜1,0 (t, x, y) where Y p˜1,0 (t, x, y)dy = 0. From the f

previous system we know that p˜1,0 (t, x, y) was completely determined. From the order ε1 in the interior, we deduce that (

∂t p1,0 + divx u1,0 + divy u2,0 = 0 ∂t u1,0 + ∇x p1,0 + ∇y p2,0 = 0

in Rd × Yf in Rd × Yf

(76)

We can integrate in y the first equation, use the fact that u2,0 · ν = 0 on ∂Ys and deduce that (p1,0 (t, x), u1,0 (t, x, y)) solves

32

Z 8 > > |Yf |∂t p1,0 + divx u1,0 = 0 > > > Yf > > > > > ∂t u1,0 + ∇x p1,0 + ∇y p2,0 = −∇x p˜1,0 (t, x, y) > > > Z > > 1 > > < divy u1,0 = −divx (u − u) |Yf | Yf > > > u1,0 · ν = 0 > > > > > p1,0 (t = 0) = 0 > > > > > > u1,0 (t = 0) = 0 > > > : y → (u1,0 , p2,0 ) is Y -periodic

in Rd in Rd × Yf (77) on ∂Ys

This system is very similar to (75). It has R R some extra forcing terms. It is important for the construction of the solution that Y divx (u − |Y1 | Y u)dy = 0. We will not detail this construction. In particular f

f

f

this system allows us to determine completely ∇y p2,0 . The rest of the construction of (pk,0 , uk,0 ) can be done by induction. For k ≥ 2, assuming that ∇y pk,0 was completely determined from the previous order, we can deduce from the order εk in the interior that we have ( ∂t pk,0 + divx uk,0 + divy uk+1,0 = 0 in Rd × Yf (78) ∂t uk,0 + ∇x pk,0 + ∇y pk+1,0 = 0 in Rd × Yf R Denoting pk,0 (t, x, y) = pk,0 (t, x) + p˜k,0 (t, x, y) where Y p˜k,0 (t, x, y)dy = 0, we know from the previous f

system R that p˜k,0 (t, x,Ry) was completely determined. Integrating in y the first equation in (78) and using that Y divy uk,0 = ∂Y uk,0 .ν = 0, we deduce that f f Z 8 > uk,0 = 0 in Rd p + div |Y |∂ x t > f k,0 > > Yf > > > > > > in Rd × Yf > ∂t uk,0 + ∇x pk,0 + ∇y pk+1,0 = −∇x p˜k,0 > > > < divy uk,0 = −divx uk−1,0 − ∂t pk−1,0 (79) > uk,0 · ν = 0 on ∂Ys > > > > > pk,0 (t = 0) = 0 > > > > > > u k,0 (t = 0) = 0 > > : y → (uk,0 , pk+1,0 ) is Y -periodic . R Notice that Y divx uk−1,0 + ∂t pk−1,0 = 0 follows from the previous order system. This process allows us f

to solve all the terms of the form (pk,0 , uk,0 ). Notice also that for now the term εβ ∆ did not contribute to these systems. Meanwhile, the order ε0 in the boundary layer gives that ubdy 0,0 (t, x, y, ξ) satisfies the following system bdy bdy 2 2 bdy ∂t ubdy 0,0 − |∇d| ∂ξξ u0,0 + ∂ξ p1,1 ∇d = 0. Taking the scalar product with ∇d, we deduce that p1,1 = 0 and bdy,tg that ubdy = ubdy solves 0,0 = u0,0

8 2 bdy ∂t ubdy − |∇d|2 ∂ξξ u =0 > > > > > bdy < u · ∇d = 0 > > > > > :

for (t, ξ) ∈ (0, ∞) × (0, ∞)

ubdy = −utg

(80) for y = Πy ∈ ∂Ys , ξ = 0

ubdy (t = 0) = 0

where we recall that the notation U tg was defined in (69). The fact that ubdy (t = 0) = 0 is consistent with the consistency condition a0 = Py a0 = 0 when y ∈ ∂Ys which implies that utg (0, y) = 0 when y ∈ ∂Ys .

33

For each x and y, system (80) is a heat equation on the half line with zero initial data and nonzero Dirichlet data. The solution can be computed explicitly (see for instance [4] and [14, 13] for other applications in boundary layers) Z t ubdy = − I(ξ, t − s)∂t utg (s, x, Πy) ds (81) 0

√ where I(ξ, t) = 2erfc(ξ/ 2t) and 8 > >
1 1 > z 2 /2 : erfc(z) = − erf(z) = √ e−˜ d˜ z 2 2π z

Notice that in (81), x and Πy are just parameters. From the order εk in the boundary layer of the second equation of (1), we get 8 bdy bdy 2 2 bdy > > ∂t uk,0 − |∇d| ∂ξξ uk,0 + ∂ξ pk+1,1 ∇d = (known) for (t, ξ) ∈ (0, ∞) × (0, ∞) > < ubdy k,0 (t = 0) = 0 > > > : ubdy for y ∈ ∂Ys , ξ = 0 k,0 = −uk,0

(82)

(83)

The normal part of (83) allows us to determine pbdy k+1,1 and the tangential part allows us to determine the tangent part of ubdy k,0 by a formula similar to (90). Step 1: The case l = 1 The order ε(β−2)/2 in the second equation and the order ε−1 in the first equation of (1) give 8 h0 bdy bdy bdy 2 2 bdy > > > ∂t u0,1 − |∇d| ∂ξξ u0,1 − ∂ξ u0,0 + ∂ξ p1,2 ∇d = 0 > h > > > < tg bdy ∂ξ ubdy 0,1 · ∇d + divy (u0,0 ) = 0 > > > ubdy > 0,1 = −u0,1 > > > : ubdy (t = 0) = 0. 0,1 Hence, ubdy 0,1 · ∇d =

∞

Z ξ

for (t, ξ) ∈ (0, ∞) × (0, ∞) (84) for y ∈ ∂Ys , ξ = 0

bdy 0 divtg y (u0,0 )dξ .

(85)

Moreover, for y ∈ ∂Ys , we have the following boundary condition Z ∞ u0,1 · ν = − divy (ubdy 0,0 )dξ 0

t

Z =

p

(86) tg 2(t − s)divtg y (∂t u (s)) ds.

0

R

∂Ys u0,1 · ν = 0 which will be necessary to solve system (88). Also, R here and in the sequel, we will often use the fact that ∂Ys divtg y (U )dσ = 0. Notice that to solve (84), we need to know u0,1 on the boundary. This means that we have first to solve the following interior problem, deduce the boundary condition utg 0,1 for y ∈ ∂Ys and then solve for

It is important to notice here that

the tangential part of ubdy 0,1 . The equation satisfied by u0,1 is 8 > ∇y p0,1 = 0, divy u0,1 = 0 > < ∂t p0,1 + divx u0,1 + divy u1,1 = 0 > > : ∂ u +∇ p +∇ p =0 t 0,1

x 0,1

y 1,1

34

in Rd × Yf in Rd × Yf in Rd × Yf

(87)

Exactly as (75) was derived, we deduce that Z 8 > > u0,1 |Yf |∂t p0,1 + divx > > > Yf > > > > > > ∂t u0,1 + ∇x p0,1 + ∇y p1,1 > > > > > divy u0,1 > > > < ∇ p y 0,1

> > > > > > > > > > > > > > > > > > > :

Z =

in Rd

u1,1 .ν ∂Yf

in Rd × Yf

=0 =0 =0 Z

∞

u0,1 · ν = − 0

(88) bdy divtg y (u0,0 )dξ

on ∂Ys

p0,1 (t = 0) = 0 u0,1 (t = 0) = 0 y → (u, p1 ) is Y -periodic

To solve (88), we need to determine the right-hand side of the first equation. From the order 0 in the bdy bdy boundary layer, we get that −∂ξ ubdy 1,1 .∇d = ∂t p0,0 + divx u0,0 and hence u1,1 .ν = −ubdy 1,1 .ν = −

∞

Z 0

divx ubdy 0,0 dξ.

This will allow to solve (88). bdy Since the normal part of ubdy 0,0 and u0,1 were already determined, taking the scalar product of (84) with ∇d, we can determine pbdy 1,2 . Now, we can solve for ubdy,tg , by looking at the part orthogonal to ∇d in (84): 0,1 8 h0 bdy,tg > 2 bdy,tg > − |∇d|2 ∂ξξ u0,1 = ∂ξ ubdy,tg > ∂t u0,1 0,0 > < h

for (t, ξ) ∈ (0, ∞) × (0, ∞) (89)

ubdy,tg (t = 0) = 0 0,1

> > > > :

ubdy,tg = −utg 0,1 0,1

for y ∈ ∂Ys , ξ = 0

This one-dimensional heat equation can also be solved explicitly (see [14, 13]): ubdy,tg =− 0,1

t

Z 0

I(ξ, t − s)∂t utg 0,1 (s, x, Πy)ds + J+ − J−

(90)

where (we recall that x and Πy are just parameters) J± =

1 2

t

Z 0

∞

Z 0

∂I h0 (ξ ± η, t − s) ∂ξ ubdy,tg (s, η)dηds. 0,0 ∂ξ h

(91)

Arguing as in the Step 0, we can continue the construction of the terms of the form uk,1 and ubdy k,1 . We do not detail that here since it is similar to the general construction that will be sketched below. Step l: The general case from l − 1 to l When β is close from 2, we need the expansion with a large number l. Here we will explain how we can keep solving for ubdy 0,l and u0,l for l = 2, 3, ..... We explain now how we can solve for these terms bdy inductively. Indeed, assuming that ubdy i,j ui,j are known for i ≥ 0 and for 0 ≤ j ≤ l − 1, we would like to construct the terms at the order l. We have from the order ε(l−1)(β−2)/2−1 in the boundary layer part of the first equation of (1) that

ubdy 0,l · ∇d =

∞

Z ξ

bdy,tg 0 divtg y (u0,l−1 )dξ

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(92)

and hence, using that p0,j is independent of y for all j and that divy u0,j = 0 for all j, we deduce that u0,l solves Z 8 > > u0,l = 0 in Rd |Y |∂ p + div t x > f 0,l > > Yf > > > > > > ∂t u0,l + ∇x p0,l + ∇y p1,l = 0 in Rd × Yf > > > > > divy u0,l+1 = 0 > > > < ∇y p0,l = 0 (93) Z ∞ > > tg bdy,tg > (u )dξ on ∂Y u · ν = − div > s 0,l y > 0,l−1 > > 0 > > > > p0,l (t = 0) = 0 > > > > > u > 0,l (t = 0) = 0 > > : y → (u0,l , p1,l ) is Y -periodic To find ubdy,tg , we use that the order εl(β−2)/2 in the boundary layer part of the second equation of 0,l (1) gives 8 bdy 2 2 bdy > ∂t ubdy for (t, ξ) ∈ (0, ∞) × (0, ∞) > 0,l − |∇d| ∂ξξ u0,l + ∂ξ p1,l+1 ∇d = (known) > < bdy (94) u0,l (t = 0) = 0 > > > bdy : u0,l = −u0,l for y ∈ ∂Ys , ξ = 0 The normal part of (94) allows to determine pbdy 1,l+1 which will be used at the step l + 1 and the tangential part will allow us to solve for ubdy 0,l . bdy Assuming that for k ≥ 1 we have constructed all the terms ubdy i,l+1 , ui,l+1 for 0 ≤ i ≤ k − 1 and that

pbdy k,l

bdy and ∇y pk,l were also determined, w e would like to construct the terms ubdy k,l , uk,l .

First notice that we have from the order ε(l−1)(β−2)/2+i in the boundary layer part of the first equation of (1) that ubdy i+1,l · ∇d =

∞

Z ξ

bdy,tg bdy bdy,tg 0 [divtg y (ui+1,l−1 ) + ∂t pi,l−1 + divx (ui,l−1 )]dξ .

(95)

This allows us to determine the normal part of ui+1,l at the boundary for each i ≥ 0. In particular it allows us to determine uk,l .ν as well as uk+1,l .ν when y ∈ ∂Yf . From the order εl(β−2)/2+k−1 in the interior, we get that (

∂t pk,l + divx uk,l + divy uk+1,l = 0

in Rd × Yf

∂t uk,l + ∇x pk,l + ∇y pk+1,l = 0

in Rd × Yf

Recall that we can write pk,l (t, x, y) = pk,l (t, x) + p˜k,l (t, x, y) where

R Yf

p˜k,l (t, x, y)dy = 0. Since ∇y pk,l

was already determined, we only need to construct pk,l (t, x). Integrating the first equation in (96) Z 8 > > |Yf |∂t pk,l + divx uk,l = F1 in Rd > > > Yf > > > > > in Rd × Yf > ∂t uk,l + ∇x pk,l + ∇y pk+1,l = F2 > > > < divy uk,l = F3 > uk,l · ν = F4 on ∂Ys > > > > > p (t = 0) = 0 > k,l > > > > > u (t = 0) = 0 k,l > > : y → (uk,l , pk,l ) is Y -periodic

36

(96)

(97)

R where the forcing terms Fa , 1 ≤ a ≤ 4 are already known and are given by F1 = − Y divy uk+1,l = f R − ∂Y uk+1,l .ν which is determined by (95), F2 = −∇x p˜k,l , F3 = −∂t pk−1,l − divx uk−1,l which is f

coming from the order k − 1 of (96) and F4 = −ubdy k,l .ν which is determined from (95). There is a compatibility R R condition between F3 and F4 for the system (97) to be well-posed, namely we need that F3 = − ∂Y F4 . This follows from the first equation of the order k − 1 of (97) since it implies that f RYf R Yf F3 = ∂Yf uk,l · ν. Hence, by induction, we conclude the construction of the asymptotic expansions. The convergence stated in theorem 3.10 can be proved by the same energy estimate as in the previous two cases and we do not detail it here.

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