Estimation of the residues and technical lemmas

The energy contained at high frequencies decays very fast. For low frequencies, we need to use assump- tion (4.51). We introduceρ >0 and we split the energy integral at a cutting threshold wheree^−ζ2t≤t^−ρ, ie. where|ζ| ≥

ρlnt t

1/2

. Note that the right-hand side tends to zero as t→+∞, enabling us to isolate only asymptotically small frequencies. Let us estimate the high frequencies energy :

W^#(t) :=

Z

|ζ|≥|^ρlnt_t |¹² 1 +ζ²² e^−2tζ2

Vˆ₀(ζ)

2

dζ

≤t^−2ρ Z

R

1 +ζ²²

Vˆ₀(ζ)

2

dζ

≤M0t^−2ρ.

(4.56)

We move on to low frequencies energy : W^[(t) :=

Z

|ζ|≤|^ρlnt^t|¹²

1 +ζ²² e^−2tζ2

Vˆ0(ζ)

2

dζ

.

ρlnt t

1 2

· sup

|ζ|≤|^ρlnt^t|¹²

Vˆ₀(ζ)

2

.

(4.57)

Thanks to assumption (4.51), a Taylor expansion atζ= 0 of ˆV0yields :

Vˆ0(ζ) ≤ 1

n!|ζ|ⁿ ∂ⁿ_ζVˆ0

∞.|ζ|ⁿ Z

R

zⁿ|V0(z)|dz.|ζ|ⁿ Z

R

(1 +z²)ⁿ⁺¹|V0(z)|²dz ¹₂

. (4.58)

Combining (4.57) with (4.58) gives :

W^[(t).Mn+2

ρlnt t

1 2+n

. (4.59)

Hence, choosingρ=¹₂+nin equation (4.56) and summing with equation (4.59) proves (4.54).Extensions

fork >0 are straightforward with Fourier computations.

Impermeability boundary condition

On ∂O, we have the impermability boundary conditionu^ε·n= 0. By identification for each power ofεin the expansion, this condition translates to the following conditions for our expansion :

u⁰·n= 0, (4.62)

v·n= 0, (4.63)

u¹+∇xθ˜¹+w+R^ε

·n= 0, (4.64)

(ρ^ε+ ˜w)·n= 0. (4.65)

By construction of the Euler trajectory u⁰, equation (4.62) is true. Since the boundary profile v is tangential, equation (4.63) is also satisfied. By construction, we also already haveu¹·n= 0. In order to be able to carry out integrations by part for the estimates of the remainder, we also would like to impose R^ε·n= 0. Thus, (4.64) yields :

∂nθ˜¹=−w_|z=0·n. (4.66)

Moreover, we see (4.65) as a definition ofu² once ˜wis known :

ρ^ε·n=−w˜_|z=0·n. (4.67)

Incompressibility condition

Similarly, the divergence-free condition on u^ε translates to :

divu⁰+n·∂_zv= 0, (4.68) div

slowv+n·∂zw= 0, (4.69) divu¹+ div∇xθ˜¹+ div

sloww+ divR^ε+n·∂_zw˜= 0, (4.70) divρ^ε+ div

sloww˜= 0. (4.71)

Here again, by construction divu⁰= divu¹=n·∂zv= 0 and we would like to work with divR^ε= 0. We also build ˜θ¹such that ∆˜θ¹= 0. Thus, we read equations (4.69), (4.70) and (4.71) as :

n·∂zw=−div

slowv, (4.72)

n·∂zw˜=−div

sloww, (4.73)

divρ^ε=−div

sloww.˜ (4.74)

The profilesw·n, ˜w·nand ˜u²are meant to catch up the unwanted divergences of previous profiles.

Slip-with-friction boundary condition

Now, we look at the slip-with-friction boundary condition on∂O: [D(u^ε)n+Au^ε]_tan= 0. To lighten the notations, let us introduce an operatorN defined as :

N(f) := [D(f)n+Af]_tan. (4.75)

Note that, for functions also depending on the fast variable,D also denotes the derivatives with respect to the slow variables in definition (4.75). Identifying the first orders of the expansion yields :

N(u⁰)−1

2∂zv= 0, (4.76)

N(v)−1

2[∂zw]_tan= 0. (4.77)

The remaining terms are left with the remainder : N(u¹) +N(∇xθ˜¹) +N(w) +N(R^ε)−1

2[∂_zw]˜ _tan+√

ε(N(ρ^ε) +N( ˜w)) = 0. (4.78)

The main boundary profilevwas built to liftN(u⁰). We choose a basic lifting forN(v) by the profilew.

Other terms are seen as a source term in the equation for the remainder :

−N(R^ε) =g^ε:=N(u¹) +N(∇xθ˜¹) +N(w) +√

εN(ρ^ε) +√

εN( ˜w). (4.79) Note that ˜wis normal and has no tangential component so we can drop it.

4.5.2 Definitions of technical profiles

Let us recall that, at this stage, the three main termsu⁰,vandu¹are defined. In this paragraph, we explain step by step how we build the following terms of the expansion.

Boundary layer pressure

Equation (4.33) only involves the tangential part of the symmetrical convective product between u⁰ andv. Hence, to compensate its normal part, we introduce as in [108] the pressureqas the unique solution vanishing asz→+∞to :

(u⁰· ∇)v+ (v· ∇)u⁰

·n=−∂zq. (4.80)

Note that the support ofqis included in the support of v. Now, we can write :

∂tv+ (u⁰· ∇)v+ (v· ∇)u⁰+u⁰_[z∂zv−∂zzv+n∂zq= 0. (4.81) Note that this pressure profile vanishes fort≥T (i.e., onceu⁰vanishes).

Second boundary correction

The first boundary conditionv generates a non vanishing slow divergence and a non vanishing normal boundary flux. The role of the profilew is to lift to unwanted terms that would be too hard to handle directly in the equation of the remainder. In order to satisfy (4.72) and (4.77), we definewby :

w(t, x, z)·n:=

Z +∞

z

div

slowv(t, x, z⁰)dz⁰, (4.82)

[w(t, x, z)]_tan:=−2 [Dx(v(t, x,0))n+Av(t, x,0)]e^−z. (4.83) Sincevis compactly supported near the boundary, so isw. Moreover,wdepends smoothly on v.

Boundary flux correction

The shape ofw·nis imposed by equation (4.72). The bad consequence is thatw·n6= 0 on∂O. Thus, we must introduce a dedicated correction∇xθ˜¹to ensure (4.66). Therefore, fort≥0, we define ˜θ¹as the solution to :

( ∆˜θ¹= 0, inO,

∂nθ˜¹=−w(t,·,0)·n, in∂O. (4.84) It can be check that the compatibility condition is indeed satisfied. This instantaneous defintion is im- portant because it guarantees that ˜u¹ := ∇xθ˜¹ will inherit decay properties proven for w. This profile solves :















∂tu˜¹+ u⁰· ∇

˜

u¹+ ˜u¹· ∇

u⁰+∇p˜¹= 0, inO fort≥0, div ˜u¹= 0 inO fort≥0,

˜

u¹·n=−w_|z=0·n in∂O fort≥0, u˜¹(0,·) = 0 inO att= 0,

(4.85)

where ˜p¹:=∂tθ˜¹−u⁰· ∇xθ˜¹.

Last instantaneous corrections Eventually, we define :

˜

w(t, x, z)·n:=

Z +∞

z

div

sloww(t, x, z⁰)dz⁰, (4.86)

which is only normal. Here again, we need to lift its normal trace. Fort≥0, we defineρ^εas the solution to :

( ∆ρ^ε=−div

sloww(t),˜ in O,

∂nρ^ε=−w(t,˜ ·,0)·n, in ∂O.

(4.87) It can be check that the compatibility condition is indeed satisfied. Thus, the profile solves the evolution equation















∂t∇xρ^ε+ u⁰· ∇

∇xρ^ε+ (∇xρ^ε· ∇)u⁰+∇µ^ε= 0, in O fort≥0, div∇xρ^ε=−div

sloww˜ in O fort≥0,

∇xρ^ε·n=−w˜_|z=0·n in ∂O fort≥0,

∇xρ^ε(0,·) = 0 in O att= 0,

(4.88)

whereµ^ε:=∂tρ^ε−u⁰· ∇xρ^ε.

4.5.3 Equation for the remainder

The equation for the remainder reads :



















∂tR^ε−ε∆R^ε+ (u^ε· ∇)R^ε+A^εR^ε+∇π^ε=F^ε−G^ε−H^ε−I^ε, in Ω fort≥0,

divR^ε= 0 in Ω fort≥0,

[D(R^ε)n+AR^ε]_tan=g^ε in ∂Ω\Γ fort≥0,

R^ε·n= 0 in ∂Ω\Γ fort≥0,

R^ε(0,·) = 0 in Ω at t= 0.

(4.89)

Before introducing the source terms, we define the following shortened notations : W^ε(t, x, z) :=w(t, x, z) +√

εw(t, x, z),˜ (4.90)

U^ε(t, x) :=u¹(t, x) + ˜u¹(t, x) +√

εu². (4.91)

Now we can introduce :

A^εR^ε:= (R^ε· ∇) u⁰+εU^ε

+ (R^ε· ∇x) √

εv+εW^ε

+ (R^ε·n) ∂zv+√

ε∂zW^ε

, (4.92)

F^ε:= [∆ϕ∂_zv+ 2∂_nzv+∂_zzW^ε] +√

ε[∆_xv+ ∆ϕ∂_zW^ε+ 2∂_nzW^ε] +ε[∆_xW^ε+ ∆U^ε], (4.93) G^ε:=u⁰·n

ϕ z∂zw+ (W^ε+U^ε)·n∂z v+√ εW^ε

+∇x

u⁰·W^ε+v² 2

+√

ε∇x[v·(W^ε+U^ε)] + ε 2∇x

(W^ε+U^ε)²

(4.94)

H^ε:=∇xq. (4.95)

I^ε:=∂tW^ε. (4.96)

Functions depending onz are evaluated atz=ϕ(x)/√

ε. Moreover, all the source terms are small when εis small. Indeed, the naturalL² scaling on the fast variable yieldsε^1/4 factors.

4.5.4 Size of the remainder

We want to prove that the remainder is sufficiently small at the final time. Since our expansion of u^ε involvesεR^ε, we need to prove here thatR^ε(T /ε,·) =o(1). As usual, we multiply equation (4.89) by R^ε and integrate by parts. Since we are controlling the equation on Γ, we can assume that our controls simulate the presence of a Navier slip-with-friction boundary condition on Γ as well (thus, on the whole of ∂Ω). We use [108, Lemma 3, page 150] to win a ε¹⁴ factor when evaluating L² norms of functions involving the fast variable. We can also perform the estimates on∂Oin the same way.

We proceed as we have done in the case of the shape operator. This basic energy method works because with have built enough technical profiles and the following estimates hold :

kA^εk_L1(L^∞)=O(1), (4.97)

kF^εk_L1(L²)=O(ε¹⁴), (4.98)

kG^εk_L1(L²)=O(ε¹⁴), (4.99)

kH^εk_L1(L²)=O(ε¹⁴), (4.100)

kI^εk_L1(L²)=O(ε¹⁴). (4.101)

Hence, we obtain :

R^ε T

ε,·

_L2

≤Cε^1/4. (4.102)

No documento DE L’UNIVERSITÉ PIERRE ET MARIE CURIE (páginas 106-110)