Improved regularity in bumpy Lipschitz domains

J. Math. Pures Appl. 113 (2018) 1–36

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Journal de Mathématiques Pures et Appliquées www.elsevier.com/locate/matpur

Improved regularity in bumpy Lipschitz domains Carlos Kenig a,1 , Christophe Prange b,a,∗,2 a b

The University of Chicago, 5734 S. University Avenue, Chicago, IL 60637, USA Université de Bordeaux, 351 cours de la Libération, 33405 Talence, France

a r t i c l e

i n f o

Article history: Received 22 September 2015 Available online 16 March 2018 MSC: 35J47 35B65 35B27 35C15 Keywords: Homogenization Elliptic systems Boundary layers Uniform Lipschitz estimates Compactness method Sobolev–Kato spaces

a b s t r a c t This paper is devoted to the proof of Lipschitz regularity, down to the microscopic scale, for solutions of an elliptic system with highly oscillating coefficients, over a highly oscillating Lipschitz boundary. The originality of this result is that it does not assume more than Lipschitz regularity on the boundary. In particular, we bypass the use of the classical regularity theory. Our Theorem, which is a significant improvement of our previous work on Lipschitz estimates in bumpy domains, should be read as an improved regularity result for an elliptic system over a Lipschitz boundary. Our progress in this direction is made possible by an estimate for a boundary layer corrector. We believe that this estimate in the Sobolev–Kato class is of independent interest. © 2018 Elsevier Masson SAS. All rights reserved.

1. Introduction This paper is devoted to the proof of Lipschitz regularity, down to the microscopic scale, for weak solutions uε = uε (x) ∈ RN of the elliptic system  ε −∇ · A(x/ε)∇uε = 0, x ∈ Dψ (0, 1), (1.1) ε ε u = 0, x ∈ Δψ (0, 1), over a highly oscillating Lipschitz boundary. Throughout this work, ψ is a Lipschitz graph, ε Dψ (0, 1) := {x ∈ (−1, 1)d−1 , εψ(x /ε) < xd < εψ(x /ε) + 1} ⊂ Rd

* Corresponding author. 1 2

E-mail addresses: [email protected] (C. Kenig), [email protected] (C. Prange). Partially supported by NSF Grants DMS-0968472 and DMS-1265249. Partially supported by NSF Grant DMS-1500893.

https://doi.org/10.1016/j.matpur.2018.03.001 0021-7824/© 2018 Elsevier Masson SAS. All rights reserved.

C. Kenig, C. Prange / J. Math. Pures Appl. 113 (2018) 1–36

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Fig. 1. The domain D ε (0, 1) and the portion of the boundary Δεψ (0, 1).

and Δεψ (0, 1) := {x ∈ (−1, 1)d−1 , xd = εψ(x /ε)} is the lower highly oscillating boundary on which homogeneous Dirichlet boundary conditions are imposed (see Fig. 1). 1.1. Statement of our results Our main theorem is the following. Theorem 1. There exists C > 0 such that for all ψ ∈ W 1,∞ (Rd−1 ), for all matrix A = A(y) = (Aαβ ij (y)) ∈ d2 ×N 2 R , elliptic with constant λ, 1-periodic and Hölder continuous with exponent ν > 0, for all 0 < ε < 1/2, for all weak solutions uε to (1.1), for all r ∈ [ε, 1/2] ˆ (−r,r)d−1

 εψ(x ˆ /ε)+r

|∇uε |2 dxd dx ≤ Crd

εψ(x /ε)

ˆ

(−1,1)d−1

 εψ(x ˆ /ε)+1

|∇uε |2 dxd dx ,

(1.2)

εψ(x /ε)

with C = C(d, N, λ, [A]C 0,ν , ψW 1,∞ ). The uniform estimate of Theorem 1 should be read as an improved regularity result. Indeed, estimate (1.2) can be seen as a Lipschitz estimate down to the microscopic scale O(ε). For an elliptic system over a slowly oscillating boundary xd = ψ(x ) for ψ merely Lipschitz, it is of course not possible to get an improved regularity estimate. What we manage to prove here is that the highly oscillating Lipschitz boundary xd = εψ(x /ε) being close in the limit to the flat boundary, system (1.1) inherits some regularity properties up to the scale O(ε) of the limit system when ε → 0. Theorem 1 represents a considerable improvement of a recent result obtained by the two authors, namely Result B and Theorem 16 in [21]. This first work dealt with uniform Lipschitz regularity over highly oscillating C 1,ν boundaries. As is classical in the proof of uniform estimates by compactness methods, we need to build (interior and boundary) correctors. Our breakthrough is made possible by estimating a boundary layer corrector v = v(y) solution to the system 

−∇ · A(y)∇v = 0, yd > ψ(y  ), v = v0 , yd = ψ(y  ),

(1.3)

C. Kenig, C. Prange / J. Math. Pures Appl. 113 (2018) 1–36

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in the Lipschitz half-space yd > ψ(y  ) with non localized Dirichlet boundary data v0 , without resorting to regularity theory. 1/2

Theorem 2. Assume ψ ∈ W 1,∞ (Rd−1 ) and v0 ∈ Huloc (Rd−1 ) i.e. sup v0 H 1/2 (ξ+(0,1)d−1 ) < ∞.

ξ∈Zd−1

Then, there exists a unique weak solution v of (1.3) such that ˆ

ˆ∞

sup

|∇v|2 dyd dy  ≤ Cv0 2H 1/2 < ∞,

(1.4)

uloc

ξ∈Zd−1

ξ+(0,1)d−1 ψ(y  )

with C = C(d, N, λ, [A]C 0,ν , ψW 1,∞ ). This estimate enables to bypass the classical regularity theory for elliptic systems over C 1,ν boundaries, which is heavily relied on in the paper [21]. Originality of our results. Let us describe two main aspects of our work: (1) the lack of regularity of the boundary, which is merely Lipschitz, (2) the lack of structure of the oscillations of the bumpy boundary. The originality of Theorem 1 lies in the fact that no smoothness of the boundary, which is just assumed to be Lipschitz, is needed for it to hold. Previous results in this direction, in particular our previous work [21], always relied on some smoothness of the boundary, typically ψ ∈ C 1,ν with ν > 0, or ψ ∈ Cω1 with ω a ´1 modulus of continuity satisfying a Dini type condition, i.e. 0 ω(t)/tdt < ∞. The difficulty of dealing with Lipschitz boundaries lies in the fact that zooming in close to the boundary does not yield any improvement of flatness. Therefore, classical regularity theory (for instance Schauder theory) is not effective. The theory for boundary value problems in Lipschitz domains, not based on regularity, is very dissimilar from the theory in C 1,ν domains. Work on boundary value problems in Lipschitz domains, in particular the development of potential theory, has started in the late 70’s and the 80’s with seminal works by Dahlberg [10,11], Dahlberg and Kenig [12] and Jerison and Kenig [18]. Recent progress toward uniform estimates for systems with oscillating coefficients has been achieved by Kenig and Shen [22], and Shen [24]. Notice in particular that estimate (1.2) of Theorem 1 implies ˆ (−1/2,1/2)d−1

 εψ(x ˆ /ε)+ε

|∇uε |2 dxd dx ≤ Cε

εψ(x /ε)

ˆ

(−1,1)d−1

 εψ(x ˆ /ε)+1

|∇uε |2 dxd dx ,

εψ(x /ε)

with a constant C uniform in ε. This is the so-called Rellich estimate (over a bumpy Lipschitz boundary), which is the keystone of the potential theory in Lipschitz domains. Pioneering work on uniform estimates in homogenization has been achieved by Avellaneda and Lin in the late 80’s [3–7]. The regularity theory for operators with highly oscillating coefficients has recently attracted a lot of attention, and important contributions have been made to relax the structure assumptions on the oscillations [8,9,15]. Our work is in a different vein. It is focused on the boundary behavior of solutions. Of course, one can flatten the boundary, and put the oscillations of the boundary into the coefficients. Here, nothing is prescribed on the boundary (except that it is Lipschitz and bounded). Notably, we do not prescribe any structure assumption on the oscillations of the boundary: ψ is neither periodic, nor quasiperiodic, nor

C. Kenig, C. Prange / J. Math. Pures Appl. 113 (2018) 1–36

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stationary ergodic. It oscillates in a completely unprescribed way. This is the main difference with the recent developments on interior estimates in homogenization, which always assume some structure on the oscillations of the coefficients. To conclude, it is remarkable that the existence of an appropriate boundary layer corrector can be proved in such generality, when the existence of bounded interior correctors in homogenization relies on some structure: see for instance [20,25,2] for almost periodic structures and [16] for random structures. Overview of the paper. In section 2 we recall several results related to Sobolev–Kato spaces, homogenization and uniform Lipschitz estimates. These results are of constant use in our work. Then the paper has two main parts. The first aim is to prove Theorem 2 about the well-posedness of the boundary layer system in a space of non localized energy over a Lipschitz boundary. The key idea is to carry out a domain decomposition. Subsequently, there are three steps. Firstly, we prove the well-posedness of the boundary layer system over a flat boundary, namely in the domain Rd+ . This is done in section 3. Secondly, we define and estimate a 1/2 Dirichlet to Neumann operator over Huloc . This key tool is introduced in section 4. Thirdly, we show that proving the well-posedness of the boundary layer system over a Lipschitz boundary boils down to analyzing a problem in a layer {ψ(y  ) < yd < 0} close to the boundary. The energy estimates for this problem are carried out in section 5. Eventually in section 6, and this is the last part of this work, we are able to prove Theorem 1 using a compactness scheme. Framework and notations. Let λ > 0 and 0 < ν < 1 be fixed in what follows. We assume that the coefficients matrix A = A(y) = (Aαβ ij (y)), with 1 ≤ α, β ≤ d and 1 ≤ i, j ≤ N is real, that A ∈ C 0,ν (Rd ),

(1.5)

that A is uniformly elliptic i.e. α β λ|ξ|2 ≤ Aαβ ij (y)ξi ξj ≤

1 2 |ξ| , λ

for all ξ = (ξiα ) ∈ RdN , y ∈ Rd

(1.6)

and periodic i.e. A(y + z) = A(y),

for all y ∈ Rd , z ∈ Zd .

(1.7)

We say that A belongs to the class Aν if A satisfies (1.5), (1.6) and (1.7). For easy reference, we summarize here the standard notations used throughout the text. For x ∈ Rd , x = (x , xd ), so that x ∈ Rd−1 denotes the d − 1 first components of the vector x. For ε > 0, r > 0, let ε Dψ (0, r) := {(x , xd ), |x | < r, εψ(x /ε) < xd < εψ(x /ε) + r} ,

Δεψ (0, r) := {(x , xd ), |x | < r, xd = εψ(x /ε)} , D0 (0, r) := {(x , xd ), |x | < r, 0 < xd < r} , Rd+ := Rd−1 × (0, ∞), Ω := {ψ(y  ) < yd < 0},

Δ0 (0, r) := {(x , 0), |x | < r} ,

Ω+ := {ψ(y  ) < yd }, Σk := (−k, k)d−1 ,

where |x | = maxi=1,... d−1 |xi |. In the whole paper, we always use the max norm of Rd−1 or Rd , so that 1 all the balls are square, not round. We sometimes write Dψ (0, r) and Δψ (0, r) in short for Dψ (0, r) and 1 Δψ (0, r); in that situation the boundary is not highly oscillating because ε = 1. Let also (u)Dψε (0,r) :=

ˆ − ε (0,r) Dψ

u=

1 ε (0, r)| |Dψ

ˆ u. ε (0,r) Dψ

C. Kenig, C. Prange / J. Math. Pures Appl. 113 (2018) 1–36

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The Lebesgue measure of a set is denoted by | · |. For a positive integer m, let also Im denote the identity matrix Mm (R). The function 1E denotes the characteristic function of a set E. The notation η usually stands for a cut-off function. Ad hoc definitions are given when needed. Unless stated otherwise, the duality product ·, · := ·, · D ,D always denotes the duality between D(Rd−1 ) = C0∞ (Rd−1 ) and D . The space of measurable functions ψ such that ψL∞ (Rd−1 ) + ∇ψL∞ (Rd−1 ) < ∞ is denoted by W 1,∞ (Rd−1 ). In the sequel, C > 0 is always a constant uniform in ε which may change from line to line. 2. Preliminaries 2.1. On Sobolev–Kato spaces s For s ≥ 0, we define the Sobolev–Kato space Huloc (Rd−1 ) of functions of non localized H s energy by

 s Huloc (Rd−1 )

:=



u∈

s Hloc (Rd ),

sup uH s (ξ+(0,1)d−1 ) < ∞ . ξ∈Zd−1

1/2

1/2

We will mainly work with Huloc . The following lemma is a useful tool to compare the Huloc norm to the H 1/2 norm of a H 1/2 (Rd−1 ) function. Lemma 3. Let η ∈ Cc∞ (Rd−1 ) and v0 ∈ Huloc (Rd−1 ). Assume that Supp η ⊂ B(0, R), for R > 0. Then, 1/2

ηv0 H 1/2 ≤ CR

d−1 2

v0 H 1/2 ,

(2.1)

uloc

with C = C(d, ηW 1,∞ ). For a proof, we refer to the proof of Lemma 2.26 in [14]. 2.2. Homogenization and weak convergence We recall the standard weak convergence result in periodic homogenization for a fixed domain Ω. As αβ usual, the constant homogenized matrix A = A ∈ MN (R) is given by αβ

A

ˆ

ˆ Aαβ (y)dy +

:= Td

Aαγ (y)∂yγ χβ (y)dy,

(2.2)

Td

where the family χ = χγ (y) ∈ MN (R), y ∈ Td , solves the cell problems ˆ −∇y · A(y)∇y χγ = ∂yα Aαγ , y ∈ Td

χγ (y)dy = 0.

and

(2.3)

Td

Theorem 4 (Weak convergence). Let Ω be a bounded Lipschitz domain in Rd and let uk ∈ H 1 (Ω) be a sequence of weak solutions to −∇ · Ak (x/εk )∇uk = fk ∈ (H 1 (Ω)) ,

C. Kenig, C. Prange / J. Math. Pures Appl. 113 (2018) 1–36

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where εk → 0 and the matrices Ak = Ak (y) ∈ L∞ satisfy (1.6) and (1.7). Assume that there exist f ∈ (H 1 (Ω)) and uk ∈ H 1 (Ω), such that fk −→ f strongly in (H 1 (Ω)) , uk → u0 strongly in L2 (Ω) and ∇uk ∇u0 weakly in L2 (Ω). Also assume that the constant matrix Ak defined by (2.2) with A replaced by Ak converges to a constant matrix A0 . Then Ak (x/εk )∇uk A0 ∇u0

weakly in L2 (Ω)

and ∇ · A0 ∇u0 = f ∈ (H 1 (Ω)) . For a proof, which relies on the classical oscillating test function argument, we refer for instance to [19, Lemma 2.1]. This is an interior convergence result, since no boundary condition is prescribed on uk . 2.3. Uniform estimates in homogenization and applications We recall here the boundary Lipschitz estimate proved by Avellaneda and Lin in [3]. Theorem 5 (Lipschitz estimate, [3, Lemma 20]). For all κ > 0, 0 < μ < 1, there exists C > 0 such that for all ψ ∈ C 1,ν (Rd−1 ) ∩ W 1,∞ (Rd−1 ), for all A ∈ Aν , for all r > 0, for all ε > 0, for all f ∈ Ld+κ (Dψ (0, r)), for all F ∈ C 0,μ (Dψ (0, r)), for all uε ∈ L∞ (Dψ (0, r)) weak solutions to 

−∇ · A(x/ε)∇uε = f + ∇ · F uε = 0,

x ∈ Dψ (0, r), x ∈ Δψ (0, r),

the following estimate holds   ∇uε L∞ (Dψ (0,r/2)) ≤ C r−1 uε L∞ (Dψ (0,r)) + r1−d/(d+κ) f Ld+κ (Dψ (0,r)) + rμ F C 0,μ (Dψ (0,r)) . (2.4) Notice that C = C(d, N, λ, κ, μ, ψW 1,∞ , [∇ψ]C 0,ν , [A]C 0,ν ). As stated in our earlier work [21], this estimate does not cover the case of highly oscillating boundaries, since the constant in (2.4) involves the C 0,ν semi-norm of ∇ψ. In this work, we rely on Theorem 5 to get large-scale pointwise estimates on the Poisson kernel P = P (y, y˜) associated to the domain Rd+ and to the operator −∇ · A(y)∇. Proposition 6. For all d ≥ 2, there exists C > 0, such that for all A ∈ Aν , we have: (1) for all y ∈ Rd+ , for all y˜ ∈ Rd−1 × {0}, we have |P (y, y˜)| ≤

Cyd , |y − y˜|d

(2.5)

|∇y P (y, y˜)| ≤

C , |y − y˜|d

(2.6)

|∇y P (y, y˜)| ≤

C . |y − y˜|d

(2.7)

(2) for all y, y˜ ∈ Rd−1 × {0}, y = y˜,

Notice that C = C(d, N, λ, [A]C 0,ν ).

C. Kenig, C. Prange / J. Math. Pures Appl. 113 (2018) 1–36

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The proof of those estimates starting from the uniform Lipschitz estimate of Theorem 5 is standard (see for instance [3]). 3. Boundary layer corrector in a flat half-space This section is devoted to the well-posedness of the boundary layer problem 

−∇ · A(y)∇v = 0, yd > 0, 1/2 d−1 v = v0 ∈ Huloc (R ), yd = 0,

(3.1)

in the flat half-space Rd+ . 1/2

Proposition 7. Assume v0 ∈ Huloc (Rd−1 ). Then, there exists a unique weak solution v of (3.1) such that ˆ

ˆ∞

sup

|∇v|2 dyd dy  ≤ Cv0 2H 1/2 < ∞,

(3.2)

uloc

ξ∈Zd−1 ξ+(0,1)d−1 0

with C = C(d, N, λ, [A]C 0,ν ). The proof is in three steps: (i) we define a function v and prove it is a weak solution to (3.1), (ii) we prove that the solution we have defined satisfies the estimate (3.2), (iii) we prove uniqueness of solutions verifying (3.2). 3.1. Existence of a weak solution Let η ∈ Cc∞ (R) a cut-off function such that η ≡ 1 on (−1, 1),

0 ≤ η ≤ 1,

η  L∞ ≤ 2.

(3.3)

Let y∗ ∈ Rd+ be fixed. Notice that η(| · −y∗ |) ∈ Cc∞ (Rd−1 ),

η(| · −y∗ |) ≡ 1 on B(y∗ , 1),

0 ≤ η(| · −y∗ |) ≤ 1 and ∇(η(| · −y∗ |))L∞ ≤ 2.

We define v(y∗ ) := v (y∗ ) + v (y∗ ), where for y ∈ Rd+ , ˆ

v (y) :=

P (y, y˜)(1 − η(|˜ y  − y∗ |))v0 (˜ y  )d˜ y,

Rd−1 ×{0}

and v = v (y) ∈ H 1 (Rd+ ) is the unique weak solution to 

yd > 0, −∇ · A(y)∇v = 0, v = η(|y  − y∗ |)v0 (y  ) ∈ H 1/2 (Rd−1 ), yd = 0,

(3.4)

C. Kenig, C. Prange / J. Math. Pures Appl. 113 (2018) 1–36

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satisfying ˆ

|∇v |2 dy  dyd ≤ Cηv0 2H 1/2 ,

(3.5)

Rd +

with C = C(d, N, λ). First of all, one has to prove that the definition of v does not depend on the choice of the cut-off η. Let η1 , η2 ∈ Cc∞ (R) be two cut-off functions satisfying (3.3). We denote by v1 (y∗ ) and v2 (y∗ ) the associated vectors defined by ˆ

P (y∗ , y˜)(1 − η1 (|˜ y  − y∗ |)v0 (˜ y  )d˜ y + v1 (y∗ ),

v1 (y∗ ) := Rd−1 ×{0}

ˆ

P (y∗ , y˜)(1 − η2 (|˜ y  − y∗ |))v0 (˜ y  )d˜ y + v2 (y∗ ).

v2 (y∗ ) := Rd−1 ×{0}

Substracting, we get ˆ

P (y∗ , y˜)(η2 (|˜ y  − y∗ |) − η1 (|˜ y  − y∗ |))v0 (˜ y  )d˜ y + v1 (y∗ ) − v2 (y∗ ).

v1 (y∗ ) − v2 (y∗ ) =

(3.6)

Rd−1 ×{0}

Now since ˆ

P (y, y˜)(η2 (|˜ y  − y∗ |) − η1 (|˜ y  − y∗ |))v0 (˜ y  )d˜ y

y −→ Rd−1 ×{0}

is the unique solution to 

−∇ · A(y)∇v = 0,

yd > 0, 

v = (η2 (|y −

y∗ |)



− η1 (|y −

y∗ |))v0 (y  )

∈H

1/2

d−1

(R

),

yd = 0,

the difference in (3.6) has to be zero, which proves that our definition of v is independent of the choice of η. It remains to prove that v = v(y) defined by (3.4) is actually a weak solution to (3.1). Let ϕ = ϕ (y  ) ∈ ∞ Cc (Rd−1 ) and ϕd = ϕd (yd ) ∈ Cc∞ ((0, ∞)). We choose η ∈ Cc∞ (R) satisfying (3.3) and such that η(| · |) ≡ 1 on Supp ϕ + B(0, 1). We aim at proving ˆ

v(y) (−∇ · A∗ (y)∇(ϕ ϕd )) dy = 0.

Rd +

This relation is clear for v . For v , by Fubini and then integration by parts ˆ

v (y) (−∇ · A∗ (y)∇(ϕ ϕd )) dy

Rd +

ˆ

ˆ

= Supp ϕ ×Supp ϕd Rd−1 ×{0}

P (y, y˜)(1 − η(˜ y ))v0 (˜ y  ) (−∇ · A∗ (y)∇ϕ ϕd ) d˜ y dy

C. Kenig, C. Prange / J. Math. Pures Appl. 113 (2018) 1–36

ˆ

ˆ

9

P (y, y˜) (−∇ · A∗ (y)∇(ϕ ϕd )) dy(1 − η(˜ y ))v0 (˜ y  )d˜ y

= Rd−1 ×{0} Supp ϕ ×Supp ϕd

ˆ

−∇ · A(y)∇P (y, y˜), ϕ ϕd (1 − η(˜ y ))v0 (˜ y  )d˜ y = 0.

= Rd−1 ×{0}

3.2. Gradient estimate Let ϕ = ϕ (y  ) ∈ Cc∞ (Rd−1 ) and ϕd = ϕd (yd ) ∈ Cc∞ ((0, ∞)). We choose R > 1 such that Supp ϕ + B(0, 1) ⊂ B(0, R). Our goal is to prove    ˆ   d−1    ∇v(y)ϕ ϕd (y)dy  ≤ CR 2 v0 H 1/2 ϕ L2 ϕd L2 , uloc    Rd +

with C = C(d, N, λ, [A]C 0,ν ). This estimate clearly implies the bound (3.2). Let η ∈ Cc∞ (R) such that (3.3) η(| · |) ≡ 1 on B(0, R)

Supp η(| · |) ⊂ B(0, 2R).

and

Combining (3.5) and the result of Lemma 3, we get ˆ

|∇v |2 dy  dyd ≤ CRd−1 v0 2H 1/2 , uloc

Rd +

with C = C(d, N, λ). It remains to estimate ˆ

ˆ1 ˆ ∇v (y)ϕ ϕd (y)dy =





0 Rd−1

Rd +

ˆ∞ ˆ

∇v (y)ϕ ϕd (y)dy dyd +



∇v (y)ϕ ϕd (y)dy  dyd .

1 Rd−1

To estimate these terms we rely on the bound (2.6): for all y ∈ Rd+ , y˜ ∈ Rd−1 × {0}, |∇y P (y, y˜)| ≤

C C = 2 , |y − y˜|d (yd + |y  − y˜ |2 )d/2

with C = C(d, N, λ, [A]C 0,ν ). We begin with two useful estimates. We have on the one hand for all y  ∈ Rd−1 such that y  + B(0, 1) ⊂ B(0, R), ˆ Rd−1

1 (1 − η(|˜ y  |))|v0 (˜ y  )|2 d˜ y =  |y − y˜ |d

ˆ Rd−1

1 (1 − η(|y  − y˜ |))|v0 (y  − y˜ )|2 d˜ y |˜ y  |d ˆ

≤ Rd−1 \B(0,1)

≤

 ξ∈Zd−1 \{0}

1 |v0 (y  − y˜ )|2 d˜ y |˜ y  |d 1 v0 2L2 ≤ Cv0 2L2 uloc uloc |ξ|d

(3.7)

C. Kenig, C. Prange / J. Math. Pures Appl. 113 (2018) 1–36

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and on the other hand for all (y  , yd ) ∈ Rd+ , ˆ

1 (1 − η(|˜ y  |))|v0 (˜ y  )|2 d˜ y ≤ 2  (yd + |y − y˜ |2 )d/2

Rd−1

ˆ Rd−1

ˆ

≤ Rd−1

≤

(yd2

+

|y 

1 |v0 (˜ y  )|2 d˜ y − y˜ |2 )d/2

1 d˜ y  v0 2L2 uloc (yd2 + |y  − y˜ |2 )d/2

C v0 2L2 . uloc yd

Using (3.7), we get  1   1 ˆ ˆ  ˆ ˆ       ∇v (y)ϕ ϕ (y)dy dy  d d =    d−1   d−1 0 R

0 R

⎛

ˆ1 ˆ 0 Rd−1



Rd−1 ×{0}

ˆ1 ˆ ≤ Cv0 L2uloc

1 − η(|˜ y |)  ⎟ d˜ y⎠  |y − y˜ |d

ˆ1 ˆ

ˆ

⎜ ⎝

⎞1/2 

Rd−1 ×{0}

1 − η(|˜ y |) ⎟ |v0 (˜ y  )|2 d˜ y⎠  |y − y˜ |d

|ϕ ϕd (y)|dy  dyd

1

|ϕ ϕd (y)|dy  dyd ≤ CR

≤ Cv0 L2uloc

       ∇y P (y, y˜)(1 − η(|˜ y |))v0 (˜ y )d˜ y ϕ ϕd (y)dy dyd   

⎛ ∞ ⎞1/2 ˆ ⎝ 1 ⎠ |ϕ ϕd (y)|dy  dyd r2

0 Rd−1

0

Rd−1 ×{0}

⎞1/2 ⎛

ˆ

⎜ ⎝

≤

ˆ

d−1 2

v0 L2uloc ϕ L2 ϕd L2 .

Rd−1

Using (3.8), we infer ∞  ˆ ˆ      ∇v (y)ϕ ϕd (y)dy dyd    d−1  1 R

ˆ∞ ˆ ≤C

⎛ ⎜ ⎝

1 Rd−1

⎞1/2

ˆ

Rd−1 ×{0}

⎛ ⎜ ⎝

1 ⎟ d˜ y ⎠ (yd2 + |y  − y˜ |2 )d/2

Rd−1 ×{0}

ˆ∞ ≤ Cv0 L2uloc

⎞1/2

ˆ (yd2

+

1 |ϕd (yd )|dyd yd

|y 

1 ⎟ |v0 (˜ y  )|2 d˜ y⎠ − y˜ |2 )d/2

ˆ

|ϕ (y  )|dy  ≤ CR

d−1 2

|ϕ ϕd (y)|dy  dyd

v0 L2uloc ϕ L2 ϕd L2 .

Rd−1

1

3.3. Uniqueness By linearity, it is enough to prove uniqueness for v = v(y) weak solution to 

−∇ · A(y)∇v = 0, v = 0,

yd > 0, yd = 0,

(3.8)

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11

such that ˆ

ˆ∞

sup

|∇v|2 dyd dy  ≤ C < ∞.

(3.9)

ξ∈Zd−1 ξ+(0,1)d−1 0

Let k be a fixed integer. A rescaled version of the Lipschitz estimate of [3] for the flat half-space reads for all n > k ˆ ˆ − |∇v|2 ≤ − |∇v|2 dyd dy  . D(0,k)

D(0,n)

Since v satisfies the bound (3.9), we get ˆ n→∞ − |∇v|2 dy ≤ Cnd−1 /nd = 1/n −→ 0. D(0,n)

Therefore, ˆ − |∇v|2 dy = 0. D(0,k)

We conclude that v = 0 on D(0, k) by using Poincaré’s inequality. 4. Estimates for a Dirichlet to Neumann operator The Dirichlet to Neumann operator DN is crucial in the proof of the well-posedness of the elliptic system in the bumpy half-space (see section 5). The key idea there is to carry out a domain decomposition. The Dirichlet to Neumann map is the tool enabling this domain decomposition. Since we are working in spaces 1/2 of infinite energy to be useful DN has to be defined on Huloc . Similar studies have been carried out in [1] (context of water-waves), [17] (2d Stokes system), [14] (3d Stokes–Coriolis system). We first define the Dirichlet to Neumann operator on H 1/2 (Rd−1 ): DN : H 1/2 (Rd−1 ) −→ D (Rd−1 ), such that for any v0 ∈ H 1/2 (Rd−1 ), for all ϕ ∈ Cc∞ (Rd−1 ), DN(v0 ), ϕ D ,D := A(y)∇v · ed , ϕ D ,D , where v is the unique weak solution to  −∇ · A(y)∇v = 0, yd > 0, 1/2 d−1 v = v0 ∈ H (R ), yd = 0.

(4.1)

Proposition 8. (1) For all ϕ ∈ Cc∞ (Rd+ ), ˆ DN(v0 ), ϕ|yd =0 D ,D = A(y)∇v · ed , ϕ|yd =0 D ,D = − Rd +

A(y)∇v · ∇ϕdy.

(4.2)

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(2) For all ϕ ∈ Cc∞ (Rd−1 ), ˆ

ˆ

DN(v0 ), ϕ|yd =0 D ,D =

A(y)∇y P (y, y˜) · ed v0 (˜ y )d˜ y ϕ(y)dy.

(4.3)

Rd−1 ×{0} Rd−1 ×{0}

For y, y˜ ∈ Rd−1 × {0}, let K(y, y˜) := A(y)∇y P (y, y˜) · ed be the kernel appearing in (4.3). Estimate (2.7) of Proposition 6 implies that |K(y, y˜)| ≤

C , |y − y˜|d

for any y, y˜ ∈ Rd−1 × {0}, y = y˜ with C = C(d, N, λ, [A]C 0,ν ). Both formulas in Proposition 8 follow from integration by parts. Because of (4.2), it is clear that for all v0 ∈ H 1/2 (Rd−1 ), for all ϕ ∈ Cc∞ (Rd−1 ), | DN(v0 ), ϕ | ≤ Cv0 H 1/2 ϕH 1/2 ,

(4.4)

with C = C(d, N, λ), so that DN(v0 ) extends as a continuous operator on H 1/2 (Rd−1 ) into H −1/2 (Rd−1 ). Another consequence of (4.2) is the following corollary. Corollary 9. For all v0 ∈ H 1/2 (Rd−1 ), ˆ DN(v0 ), v0 = −

A(y)∇v · ∇vdy ≤ 0,

Rd +

where v is the unique solution to (4.1). 1/2

Our next goal is to extend the definition of DN to v0 ∈ Huloc (Rd−1 ). We have to make sense of the duality product DN(v0 ), ϕ . As for the definition of the solution to the flat half-space problem (see section 3), the basic idea is to use a cut-off function η to split the definition between one part DN(ηv0 ), ϕ where ηv0 ∈ H 1/2 (Rd−1 ), and another part DN((1 − η)v0 ), ϕ which does not see the singularity of the kernel K(y, y˜). For R > 1, there exists η ∈ Cc∞ (R) such that 0 ≤ η ≤ 1,

η ≡ 1 on (−R, R),

Supp η ⊂ (−R − 1, R + 1),

η  L∞ ≤ 2.

Let v0 ∈ Huloc (Rd−1 ). Let R > 1 and ϕ ∈ Cc∞ (Rd−1 ) such that Supp ϕ + B(0, 1) ⊂ B(0, R). We define the action of DN(v0 ) on ϕ by 1/2

DN(v0 ), ϕ D ,D := DN(η(| · |)v0 ), ϕ H −1/2 ,H 1/2 ˆ ˆ + K(y, y˜)(1 − η(|˜ y  |))v0 (˜ y  )ϕ(y  )d˜ y dy.

(4.5)

Rd−1 ×{0} Rd−1 ×{0}

The fact that this definition does not depend on the cut-off η ∈ Cc∞ (R) follows from Proposition 8. The argument is similar to the one used in section 3.1.

C. Kenig, C. Prange / J. Math. Pures Appl. 113 (2018) 1–36

13

The first term in the right-hand side of (4.5) is estimated using (4.4) and the bound of Lemma 3 between 1/2 the H 1/2 norm of η(| · |)v0 and the Huloc norm of v0 . That yields | DN(η(| · |)v0 ), ϕ | ≤ Cη(| · |)v0 H 1/2 ϕH 1/2 ≤ CR

d−1 2

v0 H 1/2 ϕH 1/2 , uloc

with C = C(d, N, λ). We deal with the integral part in the right hand side of (4.5) in a way similar to the proof of the estimate (3.7). Using the fact that the supports of (1 − η(|y  |))v0 (y  ) on the one hand and ϕ on the other hand are disjoint, we have     ˆ ˆ        K(y, y˜)(1 − η(|˜ y |))v0 (˜ y )ϕ(y )d˜ y dy      Rd−1 ×{0} Rd−1 ×{0} ˆ ˆ 1 ≤C (1 − η(|˜ y  |))|v0 (˜ y  )||ϕ(y  )|d˜ y dy |y − y˜|d Rd−1 ×{0} Rd−1 ×{0}

⎛

ˆ

ˆ

⎜ ⎝

≤C Rd−1 ×{0}

⎛ ⎜ ⎝

Rd−1 ×{0}

ˆ

Rd−1 ×{0}

≤C Rd−1 ×{0}

≤ CR

1 ⎟ (1 − η(|˜ y  |))d˜ y⎠ |y − y˜|d ⎞1/2

1 ⎟ (1 − η(|˜ y  |))|v0 (˜ y  )|2 d˜ y⎠ |y − y˜|d

|ϕ(y  )|dy

⎞1/2 ⎛∞ ˆ 1 ⎝ dr⎠ |ϕ(y  )|dyv0 L2uloc r2

ˆ

d−1 2

⎞1/2

1

v0 L2uloc ϕL2 ,

with C = C(d, N, λ, [A]C 0,ν ). These results are put together in the following proposition. Proposition 10. (1) For v0 ∈ H 1/2 (Rd−1 ), for any ϕ ∈ Cc∞ (Rd−1 ), we have | DN(v0 ), ϕ | ≤ Cv0 H 1/2 ϕH 1/2 , with C = C(d, N, λ). 1/2 (2) For v0 ∈ Huloc (Rd−1 ), for R > 1 and any ϕ ∈ Cc∞ (Rd−1 ) such that Supp ϕ + B(0, 1) ⊂ B(0, R), we have | DN(v0 ), ϕ | ≤ CR with C = C(d, N, λ, [A]C 0,ν ).

d−1 2

v0 H 1/2 ϕH 1/2 , uloc

(4.6)

C. Kenig, C. Prange / J. Math. Pures Appl. 113 (2018) 1–36

14

Fig. 2. Splitting of the half-space Ω+ .

5. Boundary layer corrector in a bumpy half-space This section is devoted to the well-posedness of the boundary layer problem 

−∇ · A(y)∇v = 0, yd > ψ(y  ), 1/2 d−1 v = v0 ∈ Huloc (R ), yd = ψ(y  ),

(5.1)

in the bumpy half-space Ω+ := {yd > ψ(y  )}. For technical reasons, the boundary ψ ∈ W 1,∞ (Rd−1 ) is assumed to be negative, i.e. ψ(y  ) < 0 for all y  ∈ Rd−1 . We prove Theorem 2 of the introduction which asserts the existence of a unique solution v in the class ˆ

ˆ∞

sup

|∇v|2 dyd dy  < ∞.

ξ∈Zd−1 ξ+(0,1)d−1 ψ(y  )

The idea is to split the bumpy half-space into two subdomains (see Fig. 2): a flat half-space Rd+ on the one hand and a bumpy channel Ω := {ψ(y  ) < yd < 0} on the other hand. Both domains are connected by a transparent boundary condition involving the Dirichlet to Neumann operator DN defined in section 4. Therefore, solving (3.1) is equivalent to solving ⎧ ⎪ 0 > yd > ψ(y  ), ⎨ −∇ · A(y)∇v = 0, 1/2 d−1 v = v0 ∈ Huloc (R ), yd = ψ(y  ), ⎪ ⎩ A(y)∇v · e = DN(v| yd = 0. d yd =0 ),

(5.2)

This fact is stated in the following technical lemma. Lemma 11. If v is a weak solution of (5.2) in Ω such that ˆ

ˆ0

sup

|∇v|2 dyd dy  < ∞,

ξ∈Zd−1 ξ+(0,1)d−1 ψ(y  )

1/2

v|yd =0− ∈ Huloc (Rd−1 ),

(5.3)

then v˜, defined by v˜(y) := v(y) for ψ(y  ) < yd < 0 and v˜|Rd+ is the unique solution to (3.1) with boundary condition v˜|yd =0+ = v|yd =0− given by Proposition 7, is a weak solution to (5.1). Moreover, the converse is also true. Namely, if v is a weak solution to (5.1) in Ω+ such that ˆ

ˆ∞

sup ξ∈Zd−1 ξ+(0,1)d−1 ψ(y  )

then v|{ψ(y )
|∇v|2 dyd dy  < ∞,

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15

The main advantage of the domain decomposition is to make it possible to work in a channel, bounded in the vertical direction, in which one can rely on Poincaré type inequalities. Therefore our method is energy based, which makes it possible to deal with rough boundaries. We now turn to the existence of a solution of (5.2) satisfying (5.3). We lift the boundary condition v0 . There exists V0 such that ˆ

ˆ∞

sup

|V0 |2 + |∇V0 |2 dyd dy  ≤ Cv0 2H 1/2 , uloc

ξ∈Zd−1

ξ+(0,1)d−1 ψ(y  )

with C = C(d, N, ψW 1,∞ ) and such that the trace of V0 is v0 . Thus, w := v − V0 solves the system ⎧ ⎪ 0 > yd > ψ(y  ), ⎨ −∇ · A(y)∇w = ∇ · F, w = 0, yd = ψ(y  ), ⎪ ⎩ A(y)∇w · e = DN(w| d yd =0 ) + f, yd = 0,

(5.4)

where F := A(y)∇V0 , f := DN(V0 |yd =0 ) − A(y)∇V0 · ed . Notice that the source terms satisfy the following estimates: ˆ

ˆ0

sup ξ∈Zd−1 ξ+(0,1)d−1 ψ(y  )

|F |2 dyd dy  ≤ Cv0 2H 1/2 ,

(5.5)

uloc

with C = C(d, N, λ, ψW 1,∞ ) and for all ϕ ∈ Cc∞ (Rd−1 ) such that B(0, R) ⊂ Supp ϕ ⊂ B(0, 2R) for some R > 0, | f, ϕ | ≤ CR

d−1 2

v0 H 1/2 ϕH 1/2 ,

(5.6)

uloc

with C = C(d, N, λ, [A]C 0,ν , ψW 1,∞ ). There are three steps in the proof of the well-posedness of (5.1). Firstly, for n ∈ N we build approximate solutions wn = wn (y) solving ⎧ ⎪ 0 > yd > ψ(y  ), ⎨ −∇ · A(y)∇wn = ∇ · F, wn = 0, {yd = ψ(y  )} ∪ {|y  | = n}, ⎪ ⎩ A(y)∇w · e = DN(w | n d n yd =0 ) + f, yd = 0,

(5.7)

on Ω ,n := {y  ∈ (−n, n)d−1 , 0 > yd > ψ(y  )} and extend wn by 0 on Ω \ Ω ,n . We have that wn ∈ H 1 (Ω ). For n fixed, this construction is classical. Indeed, using the positivity of the Dirichlet to Neumann operator (see (4.2)) an easy energy estimate yields the bound ˆ |∇wn |2 dx  nd−1 , Ω,n

so that a Galerkin scheme makes it possible to conclude that wn exists.

16

C. Kenig, C. Prange / J. Math. Pures Appl. 113 (2018) 1–36

Fig. 3. The channel Ω , the subdomains Ω,k and Ω,y0 ,r , and the midsize box ΩT .

Secondly, we aim at getting estimates uniform in n on wn in the norm ˆ

ˆ0

sup

|∇wn |2 dyd dy  .

(5.8)

ξ∈Zd−1 ξ+(0,1)d−1 ψ(y  )

This is done by carrying out so-called Saint-Venant estimates in the bounded channel. We close this step by using a hole-filling argument. The method has been pioneered by Ladyženskaja and Solonnikov [23] for the Navier–Stokes system in a bounded channel. Here the situation is more involved because of the non local operator DN on the upper boundary. The situation here is closer to [17,13] (2d Stokes system) and [14] (3d Stokes–Coriolis system). Finally, one has to check that weak limits of wn are indeed solutions of (5.4). This step is straightforward because of the linearity of the equations. Uniqueness follows from the Saint-Venant estimate of the second step, with zero source terms. We focus on the second step, which is by far the most intricate one. We first introduce some notations for subdomains of Ω , which are displayed on Fig. 3. Let r > 0, y0 ∈ Rd−1 and Ω ,y0 ,r := {|y  − y0 | < r, 0 > yd > ψ(y  )}. Let wr ∈ H 1 (Ω ) be a weak solution to ⎧ ⎪ 0 > yd > ψ(y  ), ⎨ −∇ · A(y)∇wr = ∇ · Fr , yd = ψ(y  ), wr = 0, ⎪ ⎩ A(y)∇w · e = DN(w | r d r yd =0 ) + fr , yd = 0, such that wr = 0 on Ω \ Ω ,y0 ,r , and where Fr := F 1Ω,y ,r , 0

fr := f 1B(y0 ,r) .

(5.9)

C. Kenig, C. Prange / J. Math. Pures Appl. 113 (2018) 1–36

17

Fig. 4. Two midsize cubes T1 and T2 of volume md−1 belonging to Ck,m .

Notice that wn defined above (see (5.7)) is equal to wr solution of (5.9) for r := n and y0 = 0. The reason wr is introduced is that in section 5.1 we will need to translate the origin: wr will be a translate of wn . All estimates are carried out on wr solving the system (5.9) so that we have the uniformity of constants both in r and y0 . For k ∈ N, let Ω ,k := {y  ∈ (−k, k)d−1 , 0 > yd > ψ(y  )}. Notice that Ω ,k = Ω ,0,k with the notation above for Ω ,y0 ,r . Our goal is to estimate, ˆ |∇wr |2 dy.

Ek := Ω,k

In order to deal with the non local character of the Dirichlet to Neumann operator, we rely on a careful splitting of Rd−1 into midsize boxes of volume md−1 . ¯ even, In the following, for k, m ∈ N, k, m ≥ 1, m = 2m Σk := (−k, k)d−1 , and the set Ck,m (see Fig. 4) denotes the family of cubes T of volume md−1 contained in Rd−1 \ Σk+m−1 with vertices in Zd−1 , i.e.   Ck,m := T = ξ + (−m, ¯ m) ¯ d−1 , ξ ∈ Zd−1 and T ⊂ Rd−1 \ Σk+m−1 . Let also Cm be the family of all the cubes of volume md−1 with vertices in Zd−1   Cm := T = ξ + (−m, ¯ m) ¯ d−1 , ξ ∈ Zd−1 . Notice that for k ≥ kˆ ≥ m, ¯ Ck,m ⊂ Ck,m ⊂ Cm,m ⊂ Cm . ˆ ¯ For T ∈ Ck,m , ˆ |∇wr |2 dy,

ET := ΩT

ΩT := {y  ∈ T, 0 > yd > ψ(y  )}.

(5.10)

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Proposition 12. There exists a constant C ∗ = C ∗ (d, N, λ, [A]C 0,ν , ψW 1,∞ , v0 H 1/2 ) such that for all r > 0, uloc

y0 ∈ Rd−1 , for all k, m ∈ N, m ≥ 3 and k ≥ m/2 = m, ¯ for any weak solutions wr ∈ H 1 (Ω ) of (5.9), the following bound holds  Ek ≤ C ∗

k3d−5 kd−1 + Ek+m − Ek + 3d−3 m

 sup ET

T ∈Ck,m

.

(5.11)

Notice that C ∗ is independent of r and y0 . The crucial point for the control of the large-scale energies in (5.11) is the fact that the power 3d − 5 of k is strictly smaller that the power 3d − 3 of m. Before tackling the proof of Proposition 12, let us explain how to infer from (5.11) an a priori bound uniform in n on wn solution of (5.7). 5.1. Proof of the a priori bound in the Sobolev–Kato space: downward induction Assume that the Saint-Venant estimate of Proposition 12 has been established. We come back to its proof in section 5.2 below. Our goal is to infer from (5.11) an a priori bound ˆ |∇wn |2 dy  kd−1

Ek :=

(5.12)

Ω,k

uniform in n for wn solution of (5.7). Let us stress that wn is also a solution to (5.9) with r = n and y0 = 0. As explained above, this bound is all we need to get the existence of w solving (5.4). To show estimate (5.12), we prove (see Lemma 13 below) that the energy Em contained in one elementary midsize box of volume md−1 is bounded uniformly in m. The number m is an auxiliary parameter chosen thanks to the latitude allowed by the Saint-Venant estimate (5.11). Proving an a priori bound would be much easier if the Saint-Venant estimate (5.11) did not involve the term k3d−5 m3d−3

sup ET ,

T ∈Ck,m

whose reason for being there is the nonlocality of the Dirichlet to Neumann operator. For the sake of the explanation, assume temporarily that the Saint-Venant relation reads   Ek ≤ C ∗ k d−1 + Ek+1 − Ek ,

(5.13)

instead of (5.11). The auxiliary number which is in (5.11) to control the non local terms has no reason to appear in (5.13). For k ≥ n, Ek+1 = Ek , therefore En ≤ C ∗ nd−1 . Now, using the hole filling trick, we get  C∗  (n − 1)d−1 + C ∗ nd−1 , +1 k k   C∗ C∗ C∗ d−1 d−1 (n − k) ≤ ∗ + ... (n − 1) + C ∗ nd−1 . C +1 C∗ + 1 C∗ + 1

En−1 ≤ En−k

C∗

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Eventually,  E1 ≤

C∗ ∗ C +1

n−1

∗ d−1

C n

k n   C∗ + kd−1 C∗ + 1

 (5.14)

k=1

which is bounded uniformly in n. Let us go back to our actual estimate (5.11). The general idea of the downward induction is the same as in the simple case above. As expected yet, the non local terms make the induction trickier. Before going into the details, we fix some notation, and define the auxiliary integer m once for all. Let C ∗ be given by Proposition 12, and let k ∞   C∗ A := (C + 1) (2k − 1)d−1 < ∞, C∗ + 1

k ∞   C∗ B := (2k − 1)3d−5 < ∞. C∗ + 1

∗

k=1

(5.15)

k=1

These two numbers are the analogues of the terms appearing in the right hand side of (5.14). We now choose an integer m so that m ≥ 3,

m is even and

1 − 25−3d

B 1 > . m2 2

(5.16)

Notice that m = m(d, N, λ, [A]C 0,ν , ψW 1,∞ , v0 H 1/2 ), but is independent of r and y0 . The reason for uloc taking m even is technical; it is only used in the translation argument below. The reason for imposing the condition 1 − 25−3d

1 B > m2 2

is to be able to swallow the right hand side into the left hand side at the end of the iteration. We also take n = lm = 2lm, ¯ with l ∈ N, l ≥ 1 and let wn be the solution of (5.7). How do the non local terms make the downward induction more complicated? The trouble comes from the fact that when iterating downward the boxes Ω ,k , on which the energy Ek is computed, are always centered at 0. Nevertheless, at the end of the iteration (down to k = m) ¯ the energy Em ¯ may not be comparable in any way to sup ET ,

T ∈Cm,m ¯

which appears on the right hand side. The way out of this deadlock is to iterate downward taking into account another center ξ ∗ defined as follows. There exists T ∗ ∈ Cm such that T ∗ ⊂ Σn and ET ∗ = supT ∈Cm ET (see Fig. 5). Of course for all k ≥ m ¯ sup ET ≤

T ∈Ck,m

sup ET ≤ sup ET = ET ∗ .

T ∈Cm,m ¯

T ∈Cm

By definition, there is ξ ∗ ∈ Zd−1 for which T ∗ = ξ ∗ + (−m, ¯ m) ¯ d−1 . In order to write the iteration, it is more convenient to first center T ∗ at zero. So we now simply translate the origin. Doing so, wn∗ (y) := wn (y  + ξ ∗ , yd ) is a solution of (5.9) with y0 := −ξ ∗ , r = n and A∗ (y) := A(y  + ξ ∗ , yd ),

ψ ∗ (y  ) := ψ(y  + ξ ∗ ),

v0∗ (y  ) := v0 (y  + ξ ∗ ),

F ∗ (y) := F (y  + y ∗ , yd ) and f ∗ (y  ) := f (y  + ξ ∗ ).

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20

Fig. 5. The channel Ω seen from above: midsize cube T ∗ of volume md−1 such that ΩT ∗ concentrates most energy among the midsize cubes T ∈ Cm .

Notice that [A∗ ]C 0,μ = [A]C 0,ν ,

ψ ∗ W 1,∞ = ψW 1,∞

and v0∗ H 1/2 = v0 H 1/2 , uloc

uloc

so that wn∗ satisfies the Saint-Venant estimate (5.11) with the same constant C ∗ . Here we use the key fact that the Saint-Venant estimate (5.11) is uniform in y0 . Furthermore, Em ¯ = ET ∗ . We are now ready to state the a priori bound and to prove it by downward induction. Lemma 13. We have the following a priori bound 2−d Em Amd−1 , ¯ ≤2

where A is defined by (5.15). The Lemma is obtained by downward induction, using a hole-filling type argument. Since wn∗ is supported in Ω ,2n , we start from k sufficiently large in (5.11). For k = 2n + m ¯ = (4l + 1)m, ¯ estimate (5.11) implies ∗ E(4l+1)m ¯ d−1 , ¯ ≤ C ((4l + 1)m)

because ET = 0 for any T ∈ C(4l+1)m,m . Then, ¯ E(2(2l−1)+1)m ≤ ¯ = E(4l+1)m−m ¯

C∗ (2(2l − 1) + 1)d−1 m ¯ d−1 + C∗ + 1



C∗ C∗ + 1



C ∗ (4l + 1)d−1 m ¯ d−1 .

Let p ∈ {0, . . . 2l − 1}. We then have E(2p+1)m ¯ ≤

C∗ (2p + 1)d−1 m ¯ d−1 +1 2l−p−1 2l−p−1   C∗ C∗ d−1 d−1 + ... (4l − 1) m ¯ + C ∗ (4l + 1)d−1 m ¯ d−1 C∗ + 1 C∗ + 1   2l−p  25−3d C∗ C∗ 3d−5 3d−5 (2p + 1) + + ... (4l + 1) Em ¯. m2 C∗ + 1 C∗ + 1 C∗

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Eventually, for p = 0 Em ¯

2  C∗ C∗ d−1 m ¯ ≤ ∗ + (3m) ¯ d−1 C +1 C∗ + 1 2l−p−1 2l−p−1   C∗ C∗ d−1 d−1 + ... (4l − 1) m ¯ + C ∗ (4l + 1)d−1 m ¯ d−1 C∗ + 1 C∗ + 1    2l−p C∗ 25−3d 25−3d C∗ 3d−5 1−d + ... Em + (4l + 1) Amd−1 + BEm ¯ ≤2 ¯. 2 ∗ ∗ m C +1 C +1 m2

Therefore, Em ¯ < 2

 1 − 25−3d

B m2

 1−d Em Amd−1 , ¯ ≤2

which proves Lemma 13. Finally, ˆ

ˆ0

sup

2−d |∇wn |2 dyd dy  ≤ Em Amd−1 , ¯ ≤2

ξ∈Zd−1

ξ+(0,1)d−1 ψ(y  )

which proves the a priori bound in the norm (5.8) uniformly in n. 5.2. Proof of Proposition 12: the Saint-Venant estimate We proceed in four steps: (1) we construct a cut-off ηk with bounds uniform in k to carry out the local estimates, (2) we carry out energy estimates on system (5.9), and separate the large scales (non local effects) from the small scales, (3) we show a control of the non local terms, (4) we gather the estimates to get (5.11). Construction of a cut-off. Let η ∈ C ∞ (B(0, 1/2)) such that η ≥ 0 and ηk = ηk (y  ) be defined by 

ˆ







´ Rd

ˆ



η(y  − y˜ )d˜ y .

1[−k−1/2,k+1/2]d−1 (y − y˜ )η(˜ y )d˜ y =

ηk (y ) = Rd−1

η = 1. For all k ∈ N, let

[−k−1/2,k+1/2]d−1

For all k ∈ N, we have the following properties: ηk ≡ 1 on [−k, k]d−1 ,

Supp ηk ⊂ [−k − 1, k + 1]d−1 ,

and most importantly, we have the control ∇ηk L∞ ≤ ∇ηL1 uniformly in k.

ηk ∈ Cc∞ (Rd−1 )

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22

Energy estimate. Testing the system (5.9) against ηk2 wr we get ˆ

ˆ ηk2 A(y)∇wr · ∇wr dy = − Ω

2ηk A(y)∇wr · ∇ηk wr dy

Ω

+ ∇ · Fr , ηk2 wr + fr , ηk2 wr + DN(wr |yd =0 ), ηk2 wr .

(5.17)

By ellipticity, we have ˆ

ˆ ηk2 |∇wr |2 dy

λ

≤

Ω

ηk2 A(y)∇wr · ∇wr dy. Ω

Using that ηk2 wr vanishes on the lower oscillating boundary, the fundamental theorem of calculus yields a Poincaré type inequality ηk2 wr L2 (Ω ) ≤ C∂yd (ηk2 wr )L2 (Ω ) ,

(5.18)

where the constant C only depends on the height of the channel Ω , and therefore not on k. The following estimate (or variations of it) is of constant use: by the trace theorem and the Poincaré inequality (5.18) ⎛ ⎜ ⎝

⎞1/2

ˆ

⎟ ηk4 |wr (y  , 0)|2 dy  ⎠

Σk+1

  ≤ ηk2 wr H 1/2 ≤ C ηk2 wr L2 (Ω ) + ∇(ηk2 wr )L2 (Ω )   ≤ C ∂yd (ηk2 wr )L2 (Ω ) + ∇(ηk2 wr )L2 (Ω ) ≤ C∇(ηk2 wr )L2 (Ω ) ⎞1/2 ⎛ ˆ ≤ C(Ek+1 − Ek )1/2 + C  ⎝ ηk4 |∇wr |2 dy ⎠ ,

(5.19)

Ω

with C = C(d, ψW 1,∞ , ηL1 ) and C  = C  (d). We now estimate every term on the right hand side of (5.17). We have,   ⎞1/2 ⎛ ⎞1/2 ⎛ ˆ  ˆ ˆ   2  2ηk A(y)∇wr · ∇ηk wr dy  ≤ ⎝ ηk2 |∇wr |2 dy ⎠ ⎝ |∇ηk |2 |wr |2 dy ⎠   λ   Ω

⎛ ≤C⎝

Ω

ˆ

⎞1/2 ηk2 |∇wr |2 dy ⎠

Ω

(Ek+1 − Ek )1/2 ,

Ω

with C = C(λ, ηL1 ). We also have, | ∇ · Fr , ηk2 wr | = | ∇ · F, ηk2 wr | = | F, ∇(ηk2 wr ) | ≤ Ck

d−1 2

(Ek+1 − Ek )1/2 + C  k

⎛ d−1 2

⎝

ˆ

Ω

⎞1/2 ηk4 |∇wr |2 dy ⎠

,

C. Kenig, C. Prange / J. Math. Pures Appl. 113 (2018) 1–36

23

where C = C(v0 H 1/2 , ∇ηL1 ) and C  = C  (v0 H 1/2 ), and by the trace theorem and Poincaré inequality uloc

uloc

| fr , ηk2 wr | = | f, ηk2 wr | ≤ Ck

d−1 2

ηk2 wr H 1/2 ≤ Ck

d−1 2

∇(ηk2 wr )L2 ⎞1/2 ⎛ ˆ d−1 d−1 ≤ Ck 2 (Ek+1 − Ek )1/2 + C  k 2 ⎝ ηk4 |∇wr |2 dy ⎠ , Ω

with C = C(d, ψW 1,∞ , v0 H 1/2 , ∇ηL1 ) and C  = C  (v0 H 1/2 ). We have now to tackle the non local uloc uloc term involving the Dirichlet to Neumann operator. We split this term into 2 DN(wr |yd =0 ), ηk2 wr = DN((1 − ηk+m−1 )wr |yd =0 ), ηk2 wr

2 − ηk2 )wr |yd =0 ), ηk2 wr + DN(ηk2 wr ), ηk2 wr . + DN((ηk+m−1

By Corollary 9, DN(ηk2 wr ), ηk2 wr ≤ 0. Relying on Proposition 10 and on estimate (4.6), we get 2 | DN((ηk+m−1 − ηk2 )wr |yd =0 ), ηk2 wr | ≤ Ck

⎛ ≤ C(Ek+m − Ek )1/2 ⎝

ˆ

d−1 2

2 (ηk+m−1 − ηk2 )wr |yd =0 H 1/2 ηk2 wr H 1/2 uloc

⎞1/2 ηk4 |∇wr |2 dy ⎠

+ C(Ek+m − Ek )1/2 (Ek+1 − Ek )1/2 ,

Ω

with C = C(d, N, λ, [A]C 0,ν , ψW 1,∞ , ηL1 ). Notice that the bound (4.4) for the Dirichlet to Neumann operator in H 1/2 here is actually enough, since wr is compactly supported. However, when dealing with solutions not compactly supported, as for the uniqueness proof in section 5.3, we have to use the result of Proposition 10. Control of the non local term. Lemma 14. For all m ≥ 3, all k ≥ m ¯ = m/2, we have ˆ Σk+1

⎛ ⎝

ˆ Rd−1

⎞2 1 k3d−5 2  ⎠  (1 − η )|w (˜ y , 0)|d˜ y dy ≤ C r k+m−1 |y  − y˜ |d m3d−3

sup ET ,

T ∈Ck,m

where C = C(d). Let y  ∈ Σk+1 be fixed. We have ˆ Rd−1

=

|y 

1 2 (1 − ηk+m−1 )|wr (˜ y  , 0)|d˜ y − y˜ |d

ˆ ∞  j=1

Rd−1

|y 

1 2 (η 2 − ηk+j(m−1) )|wr (˜ y  , 0)|d˜ y − y˜ |d k+(j+1)(m−1)

(5.20)

C. Kenig, C. Prange / J. Math. Pures Appl. 113 (2018) 1–36

24

≤

j=1

=

ˆ

∞ 

Σk+(j+1)(m−1)+1 \Σk+j(m−1)

∞ 



ˆ

j=1 T ∈Ck,j,m T

|y 

1 |wr (˜ y  , 0)|d˜ y |y  − y˜ |d

1 |wr (˜ y  , 0)|d˜ y , − y˜ |d

where Ck,j,m is a family of disjoint cubes T = ξ + (−m, ¯ m) ¯ d−1 such that T ⊂ Σk+(j+1)(m−1)+1 \ Σk+j(m−1) and



T ∈Ck,j,m

T = Σk+(j+1)(m−1)+1 \ Σk+j(m−1) .

For all T ∈ Ck,j,m , by Cauchy–Schwarz, trace theorem and Poincaré inequality ˆ T

⎛

⎞1/2 ⎛

ˆ

1 |wr (˜ y  , 0)|d˜ y ≤ ⎝ |y  − y˜ |d

1 d˜ y ⎠ |y  − y˜ |2d

T

⎛

≤C⎝

ˆ

T

⎛ ≤C⎝

ˆ

T

⎞1/2

ˆ

⎝

|wr (˜ y  , 0)|2 d˜ y ⎠

T

⎞1/2 ⎛ 1 d˜ y ⎠ |y  − y˜ |2d

⎝

⎞1/2 |∇wr |2 d˜ y ⎠

ΩT

⎞1/2 

1 d˜ y ⎠ |y  − y˜ |2d

ˆ

1/2 sup

T ∈Ck,j,m

ET

,

where ΩT and ET are defined in (5.10). Notice that the constant C in the last inequality only depends on d and on ψW 1,∞ . Moreover, for any T ∈ Ck,j,m , ⎛

⎞1/2 d−1 m 2 1 ⎠ d˜ y ≤ , |y  − y˜ |2d (k + j(m − 1) − |y  |)d

ˆ

⎝

T

and the number of elements of Ck,j,m is bounded by   Σk+(j+1)(m−1)+1 \ Σk+j(m−1) 

#Ck,j,m =

md−1



(k + j(m − 1))d−2 . md−2

Therefore, ˆ Rd−1

≤C

|y 

1 2 (1 − ηk+m−1 )|wr (˜ y  , 0)|d˜ y − y˜ |d



1/2 sup

T ∈Ck,j,m

ET 1/2

sup

T ∈Ck,j,m

ET

∞  j=1

 ≤C



1 m

1/2 sup

T ∈Ck,j,m

d−1

j=1 T ∈Ck,j,m

 ≤C

∞ 

ET

d−3 2

m 2 (k + j(m − 1) − |y  |)d

(k + j(m − 1))d−2 (k + j(m − 1) − |y  |)d

(k + m − 1)d−2 m

d−1 2

(k + m − 1 − |y  |)d−1

,

C. Kenig, C. Prange / J. Math. Pures Appl. 113 (2018) 1–36

25

with C = C(d). Eventually, we get for m ≥ 3 ⎛

ˆ

ˆ

⎝

Σk+1



≤C

Rd−1

⎞2 1 2 (1 − ηk+m−1 )|wr (˜ y  , 0)|d˜ y  ⎠ dy  |y  − y˜ |d 

sup

T ∈Ck,j,m

ET

 ≤C

 sup

T ∈Ck,j,m

ET

ˆ

(k + m − 1)2d−4 md−1

Σk+1

1 dy  (k + m − 1 − |y  |)2d−2

(k + m − 1)2d−4 (k + 1)d−1 k3d−5 ≤ C md−1 (m − 2)2d−2 m3d−3

sup

T ∈Ck,j,m

ET ,

with C = C(d), the last inequality being only true on condition that k ≥ m/2 = m. ¯ This proves Lemma 14. 2 y  , 0) and ηk2 wr (y  , 0) In particular, by the definition of DN in (4.5), by the fact that (1 − ηk+m−1 )wr (˜ have disjoint support, by estimate (5.20) and by the bound (5.19) we get 2 | DN((1 − ηk+m−1 )wr |yd =0 ), ηk2 wr |     ˆ ˆ   1 2   2      (1 − ηk+m−1 (˜ y ))|wr (˜ y , 0)|ηk (y )|wr (y , 0)|d˜ y dy  ≤C −y  |d |y ˜   d−1 d−1 R

⎛ ⎜ ≤C⎝

R

ˆ

⎞1/2 ⎛ ⎟ ηk4 |wr (y  , 0)|2 dy  ⎠

⎜ ⎝

Σk+1

≤C

3d−5 2

m

3d−3 2

⎜ ⎝ ⎡

≤C

k

3d−5 2

m

3d−3 2

⎛ ⎝

⎞1/2

ˆ

⎟ ηk4 |wr (y  , 0)|2 dy  ⎠

Σk+1

⎛

⎢ 1/2 ⎣(Ek+1 − Ek ) + ⎝

ˆ

ˆ

⎞1/2

⎞2

1

Rd−1

Σk+1

⎛ k

ˆ

2 − ηk+m−1 (˜ y ) |wr (y  , 0)|d˜ y ⎠ |y  − y˜ |d



⎟ dy  ⎠

,

 12 sup

T ∈Ck,j,m

ET

⎞1/2 ⎤  ⎥ sup ηk4 |∇w|2 dy ⎠ ⎦

 12

T ∈Ck,j,m

Ω

ET

,

with C = C(d, N, λ, [A]C 0,ν , ψW 1,∞ ). End of the proof of the Saint-Venant estimate. Combining all our bounds and using Ek+1 − Ek ≤ Ek+m − Ek ,

ηk4 ≤ C(ηL∞ )ηk2

whenever possible, we get from (5.17) the following estimate ˆ λ

⎛ ηk2 |∇wr |2 dy ≤ C ⎝

Ω

⎞1/2

ˆ

ηk2 |∇wr |2 dy ⎠

Ω

+ Ck

⎛

d−1 2

⎝

ˆ

Ω

⎡

+C

k

3d−5 2

m

3d−3 2

(Ek+m − Ek )1/2 + Ck ⎞1/2

ηk2 |∇wr |2 dy ⎠

d−1 2

(Ek+m − Ek )1/2 + C(Ek+m − Ek ) ⎛

+ C(Ek+m − Ek )1/2 ⎝

ˆ

⎞1/2 ηk2 |∇wr |2 dy ⎠

Ω

⎞1/2 ⎤  ˆ ⎢ ⎥ 1/2 2 2 sup ⎣(Ek+1 − Ek ) + ⎝ ηk |∇w| dy ⎠ ⎦ ⎛

Ω

T ∈Ck,j,m

1/2 ET

,

26

C. Kenig, C. Prange / J. Math. Pures Appl. 113 (2018) 1–36

with C = C(d, N, λ, [A]C 0,ν , v0 H 1/2 , ψW 1,∞ ). Swallowing every term of the type uloc

ˆ ηk2 |∇w|2 dy Ω

in the left hand side, we end up with the Saint-Venant estimate (5.11). This concludes the proof of Proposition 12. 5.3. End of the proof of Theorem 2: uniqueness Extracting subsequences using a classical diagonal argument and passing to the limit in the weak formulation of (5.7) relying on the continuity of the Dirichlet to Neumann map asserted in estimate (4.4) yields the existence of a weak solution w to the system (5.4). In addition, the weak solution satisfies the bound ˆ

ˆ∞

|∇w|2 dyd dy  ≤ 22−d Amd−1 < ∞.

sup

(5.21)

ξ∈Zd−1

ξ+(0,1)d−1 ψ(y  )

Let us turn to the uniqueness of the solution to (5.4) satisfying the bound (5.21). By linearity of the 1 problem, it is enough to prove the uniqueness for zero source terms. Assume w ∈ Hloc (Ω ) is a weak solution to (5.4) with f = F = 0 satisfying ˆ

ˆ0

sup

|∇w|2 dyd dy  ≤ C0 < ∞.

(5.22)

ξ∈Zd−1

ξ+(0,1)d−1 ψ(y  )

Repeating the estimates leading to Proposition 12 (see section 5.2), we infer that for the same constant C ∗ appearing in the Saint-Venant estimate (5.11) and for m defined by (5.16), for k ∈ N, k ≥ m/2 = m, ¯  Ek ≤ C

∗

k3d−5 Ek+m − Ek + 3d−3 m

 sup ET

T ∈Ck,m

.

(5.23)

The fact that w, unlike wn , does not vanish outside Ω ,n does not lead to any difference in the proof of this estimate. Since sup ET < ∞,

T ∈Cm

for any ε, there exists Tε∗ ∈ Cm such that sup ET − ε ≤ ETε∗ ≤ sup ET .

T ∈Cm

T ∈Cm

(5.24)

Again, Tε∗ := ξε∗ + (−m, ¯ m) ¯ d−1 for ξε∗ ∈ Zd−1 , and we can translate Tε∗ so that it is centered at the origin as has been done in section 5.1. Estimate (5.23) still holds. For any n ∈ N, En ≤ C0 nd−1 where C0 is defined by (5.22). The idea is now to carry out a downward iteration. For any n = (2l + 1)m ¯ with l ∈ N, l ≥ 1 fixed, for p ∈ {1, . . . l − 1} one can show that

C. Kenig, C. Prange / J. Math. Pures Appl. 113 (2018) 1–36



E(2p+1)m ¯

27

2 l−p   C∗ C∗ + ... En C∗ + 1 C∗ + 1   l−p  ∗ 25−3d C∗ C (2p + 1)3d−5 + . . . + (2l + 1)3d−5 sup ET m2 C∗ + 1 C∗ + 1 T ∈Cm   l−p+1 C∗ C∗ + 1 ≤ C0 ∗ nd−1 ∗ 2C + 1 C + 1   l−p  ∗ 25−3d C∗ C (2p + 1)3d−5 + . . . + (2l + 1)3d−5 sup ET . m2 C∗ + 1 C∗ + 1 T ∈Cm C∗ + ≤ C∗ + 1



Thus, Em ¯ ≤ C0

C∗ + 1 2C ∗ + 1

C∗ + 1 ≤ C0 ∗ 2C + 1

 

C∗ ∗ C +1 C∗ C∗ + 1

l (2l + 1)d−1 m ¯ d−1 + 2l+1

25−3d B sup ET m2 T ∈Cm

(2l + 1)d−1 m ¯ d−1 +

25−3d B(Em ¯ + ε). m2

From this we infer using (5.16) that Em ¯

C∗ + 1 ≤ 2C0 ∗ 2C + 1



C∗ C∗ + 1

2l+1 (2l + 1)d−1 m ¯ d−1 +

6−3d 26−3d l→∞ 2 Bε −→ Bε. m2 m2

Therefore, from equation (5.24)   26−3d sup ET ≤ 1 + B ε, m2 T ∈Cm which eventually leads to supT ∈Cm ET = 0, or in other words w = 0. Combining this existence and uniqueness result for the system (5.4) in the bumpy channel Ω with Lemma 11 and Proposition 7 about the well-posedness in the flat half-space finishes the proof of Theorem 2. 6. Improved regularity over Lipschitz boundaries The goal in this section is to prove Theorem 1 of the introduction. Let us recall the result we prove in the following proposition. Proposition 15. For all ν > 0, γ > 0, there exist C > 0 and ε0 > 0 such that for all ψ ∈ W 1,∞ (Rd−1 ), −1 < ψ < 0 and ∇ψL∞ ≤ γ, for all A ∈ Aν , for all 0 < ε < (1/2)ε0 , for all weak solutions uε to (1.1), for all r ∈ [ε/ε0 , 1/2] ˆ − ε (0,r) Dψ

ˆ −

|∇uε |2 ≤ C

|∇uε |2 ,

ε (0,1) Dψ

or equivalently, ˆ − ε (0,r) Dψ

with C = C(d, N, λ, ν, γ, [A]C 0,ν ).

|uε |2 ≤ Cr2

ˆ −

ε (0,1) Dψ

|uε |2 ,

(6.1)

C. Kenig, C. Prange / J. Math. Pures Appl. 113 (2018) 1–36

28

We rely on a compactness argument inspired by the pioneering work of Avellaneda and Lin [3,6], and our recent work [21]. The proof is in two steps. Firstly, we carry out the compactness argument. Secondly, we iterate the estimate obtained in the first step, to get an estimate down to the microscopic scale O(ε). A key step in the proof of boundary Lipschitz estimates is to estimate boundary layer correctors, which is done by combining the classical Lipschitz estimate with a uniform Hölder estimate, as in [3, Lemma 17] or [21, Lemma 10]. Here, we are able to relax the regularity assumption on ψ. This progress is enabled by our new estimate (1.4) for the boundary layer corrector, which holds for Lipschitz boundaries ψ. We begin with an estimate which is of constant use in this part of our work. Take ψ ∈ W 1,∞ (Rd−1 ) and A ∈ Aν . By Cacciopoli’s inequality, there exists C > 0 such that for all ε > 0, for all weak solutions uε to 

ε −∇ · A(x/ε)∇uε = 0, x ∈ Dψ (0, 1), uε = 0, x ∈ Δεψ (0, 1),

(6.2)

for all 0 < θ < 1,   ⎛ ⎞1/2  ˆ  ˆ     ε 2⎟ (∂x uε )D (0,θ)  =  − ∂x uε  ≤ ⎜ ⎝ − |∂xd u | ⎠ d d ψ   Dψ (0,θ)  Dψ (0,θ) ⎛ ≤

C0 d/2 θ (1 −

θ)

⎜ ⎝

ˆ −

⎞1/2 ⎟ |uε |2 ⎠

(6.3)

.

Dψ (0,1)

Notice that C0 in (6.3) only depends on λ. Proposition 15 is a consequence of the two following lemmas. The first one contains the compactness argument. The second one is the iteration lemma. In order to alleviate the statement of the following lemma, the definition of the boundary layer v is given straight after the lemma. Lemma 16. For all ν > 0, γ > 0, there exists θ > 0, 0 < μ < 1, ε0 > 0, such that for all ψ ∈ W 1,∞ (Rd−1 ), −1 < ψ < 0 and ∇ψL∞ ≤ γ, for all A ∈ Aν , for all 0 < ε < ε0 , for all weak solutions uε to (6.2) we have ˆ −

|uε |2 ≤ 1

ε (0,1) Dψ

implies ˆ −

 " #2  ε u (x) − (∂xd uε )Dψε (0,θ) xd + εχd (x/ε) + εv(x/ε)  ≤ θ2+2μ .

ε (0,θ) Dψ

The boundary layer v = v(y) is the unique solution given by Theorem 2 to the system 

−∇ · A(y)∇v = 0, v = −yd − χd (y),

yd > ψ(y  ), yd = ψ(y  ).

(6.4)

The estimate of Theorem 2 implies ˆ

ˆ∞

sup ξ∈Zd−1 ξ+(0,1)d−1 ψ(y  )

  |∇v|2 dyd dy  ≤ C ψH 1/2 (Rd−1 ) + χ(·, ψ(·))H 1/2 (Rd−1 ) , uloc

uloc

C. Kenig, C. Prange / J. Math. Pures Appl. 113 (2018) 1–36

29

with C = C(d, N, λ, [A]C 0,ν , ψW 1,∞ ). Now, by Sobolev injection W 1,∞ (Rd−1 ) → H 1/2 (Rd−1 ) ψH 1/2 (Rd−1 ) ≤ CψW 1,∞ (Rd−1 ) , uloc

with C = C(d) and by classical interior Lipschitz regularity χ(·, ψ(·))H 1/2 ≤ Cχ(·, ψ(·))W 1,∞ (Rd−1 ) ≤ CχW 1,∞ (Rd ) ≤ C, uloc

with in the last inequality C = C(d, N, λ, [A]C 0,ν ). Eventually, ˆ

ˆ∞

sup

|∇v|2 dyd dy  ≤ C,

(6.5)

ξ∈Zd−1 ξ+(0,1)d−1 ψ(y  )

with C = C(d, N, λ, [A]C 0,ν , ψW 1,∞ ) uniform in ε. The interior corrector χd as well as the boundary layer corrector v are crucial in the iteration procedure, which is the second step of the method. Lemma 17. Let θ, ε0 and γ be given as in Lemma 16. For all ψ ∈ W 1,∞ (Rd−1 ), −1 < ψ < 0 and ∇ψL∞ ≤ γ, for all A ∈ Aν , for all k ∈ N, k > 0, for all 0 < ε < θk−1 ε0 , for all weak solutions uε to (6.2) there exists aεk ∈ RN satisfying |aεk | ≤ C0

1 + θμ + . . . θμ(k−1) , θd/2 (1 − θ)

such that ˆ −

|uε |2 ≤ 1

ε (0,1) Dψ

implies ˆ −

 ε " # u (x) − aεk xd + εχd (x/ε) + εv(x/ε) 2 ≤ θ(2+2μ)k ,

(6.6)

ε (0,θ k ) Dψ

where v = v(y) is the solution, given by Theorem 2, to the boundary layer system (6.4). The condition ε < θk−1 ε0 can be seen as giving a lower bound on the scales θk for which one can prove the regularity estimate: θk−1 > ε/ε0 . In that perspective, estimate (6.6) is an improved C 1,μ estimate down to the microscale ε/ε0 . For fixed 0 < ε/ε0 < 1/2 and r ∈ [ε/ε0 , 1/2], there exists k ∈ N such that θk+1 < r ≤ θk . We aim at estimating ˆ − ε (0,r) Dψ

|uε (x)|2

C. Kenig, C. Prange / J. Math. Pures Appl. 113 (2018) 1–36

30

using the bound (6.6). We have ⎛ ⎜ ⎝

⎞1/2

ˆ −

⎛

⎟ |uε (x)|2 ⎠

⎜ ≤⎝

ε (0,r) Dψ

⎛ ⎜ ≤⎝

⎞1/2

ˆ −

⎟ |uε (x)|2 ⎠

ε (0,θ k ) Dψ

⎞1/2

ˆ −

 ε # " u (x) − aεk xd + εχd (x/ε) + εv(x/ε) 2 ⎟ ⎠

(6.7)

ε (0,θ k ) Dψ

⎧⎛ ⎞1/2 ⎛ ⎞1/2 ⎛ ⎞1/2 ⎫ ⎪ ⎪ ⎪ ⎪ ˆ ˆ ˆ ⎨ ⎬ ⎜ ⎜ ⎜ ε 2⎟ d 2⎟ 2⎟ + |ak | ⎝ − |xd | ⎠ + ⎝ − |εχ (x/ε)| ⎠ + ⎝ − |εv(x/ε)| ⎠ . ⎪ ⎪ ⎪ ⎪ ⎩ Dε (0,θk ) ⎭ ε ε k k D (0,θ ) D (0,θ ) ψ

ψ

ψ

Let us focus on the term involving the boundary layer. Let η = η(yd ) ∈ Cc∞ (R) be a cut-off such that η ≡ 1 on (−1, 1) and Supp η ⊂ (−2, 2). The triangle inequality yields ⎛

⎞1/2

ˆ −

⎜ ⎝

⎟ |εv(x/ε)|2 ⎠

⎛

ˆ −

⎜ ≤⎝

ε (0,θ k ) Dψ

⎞1/2 ⎟ |εv(x/ε) − (xd + εχd (x/ε))η(xd /ε)|2 ⎠

ε (0,θ k ) Dψ

⎛ ⎜ +⎝

ˆ −

⎞1/2 ⎟ |(xd + εχd (x/ε))η(xd /ε)|2 ⎠

.

ε (0,θ k ) Dψ

Poincaré’s inequality implies ⎛ ⎜ ⎝

⎞1/2

ˆ −

⎟ |εv(x/ε) − (xd + εχd (x/ε))η(xd /ε)|2 ⎠

ε (0,θ k ) Dψ

⎛

⎞1/2

ˆ −

⎜ ≤ θk ⎝

   ∇ εv(x/ε) − (xd + εχd (x/ε))η(xd /ε) 2 ⎟ ⎠

ε (0,θ k ) Dψ

⎛

⎞1/2

ˆ −

⎜ ≤ θk ⎝

⎟ |∇v(x/ε)|2 ⎠

⎛ ⎜ + (1 + ∇χ2L∞ )θk ⎝

ε (0,θ k ) Dψ

ε (0,θ k ) Dψ

⎛ k

+

ˆ −

θ ⎜ ⎝ ε

⎞1/2

ˆ −

⎟ |(xd + εχd (x/ε))η  (xd /ε)|2 ⎠

ε (0,θ k ) Dψ

Estimate (6.5) now yields ˆ − ε (0,θ k ) Dψ

|∇v(x/ε)|2 ≤ Cεθ−k ,

.

⎞1/2 ⎟ |η(xd /ε)|2 ⎠

C. Kenig, C. Prange / J. Math. Pures Appl. 113 (2018) 1–36

31

so that eventually using ε/ε0 ≤ r ≤ θk , ⎛ ⎜ ⎝

ˆ −

⎞1/2 ⎟ |εv(x/ε) − (xd + εχd (x/ε))η(xd /ε)|2 ⎠

  θk ≤ C ε1/2 θk/2 + θk + (ε + ε) ≤ Cθk ε

ε (0,θ k ) Dψ

with C = C(d, N, λ, [A]C 0,ν , ψW 1,∞ ). It follows from (6.7) and (6.6) that ⎛ ⎜ ⎝

⎞1/2

ˆ −

⎟ |uε (x)|2 ⎠

≤ θ(1+μ)k + Cθk ≤ Cθk ≤ Cr,

ε (0,r) Dψ

which is the estimate of Proposition 15. 6.1. Proof of Lemma 16 Let 0 < θ < 1/8 and u0 ∈ H 1 (D0 (0, 1/4)) be a weak solution of 

−∇ · A0 ∇u0 = 0, u0 = 0,

x ∈ D0 (0, 1/4), x ∈ Δ0 (0, 1/4),

(6.8)

such that ˆ −

|u0 |2 ≤ 4d .

D0 (0,1/4)

The classical regularity theory yields u0 ∈ C 2 (D0 (0, 1/8)). Using that for all x ∈ D0 (0, θ) ' ( u0 (x) − ∂xd u0

0,θ

' ( xd = u0 (x) − u0 (x , 0) − ∂xd u0 =

1 |D0 (0, θ)|

ˆ1

ˆ



0,θ

xd

 ∂xd u0 (x , txd ) − ∂xd u0 (y) xd dxdt,

0 D0 (0,θ)

we get ˆ −

 2  0  ) 4, u (x) − (∂xd u0 )D0 (0,θ) xd  ≤ Cθ

(6.9)

D0 (0,θ)

) = C(d, ) N, λ). Fix 0 < μ < 1. Choose 0 < θ < 1/8 sufficiently small such that where C ) 4. θ2+2μ > Cθ

(6.10)

The rest of the proof is by contradiction. Fix γ > 0. Assume that for all k ∈ N, there exists ψk ∈ W 1,∞ (Rd−1 ), −1 < ψ < 0 and ψk L∞ ≤ γ, there exists Ak ∈ Aν , there exists 0 < εk < 1/k, there exists uεkk solving

(6.11)

C. Kenig, C. Prange / J. Math. Pures Appl. 113 (2018) 1–36

32



−∇ · Ak (x/εk )∇uεkk = 0, uεkk

= 0,

εk x ∈ Dψ (0, 1), k

x ∈ Δεψkk (0, 1),

such that ˆ −

|uεkk |2 ≤ 1

(6.12)

 #2 "  εk uk (x) − (∂xd uεkk )Dψεk (0,θ) xd + εk χdk (x/εk ) + εk vk (x/εk )  > θ2+2μ .

(6.13)

ε Dψk (0,1) k

and ˆ −

k

ε

Dψk (0,θ) k

Notice that χdk is the cell corrector associated to the operator −∇ · Ak (y)∇ and vk is the boundary layer corrector associated to −∇ · Ak (y)∇ and to the domain yd > ψk (y  ). First of all, for technical reasons, let us extend uεkk by zero below the boundary, on {x ∈ (−1, 1)d−1 , xd ≤ εk ψk (x /εk )}. The extended functions are still denoted the same, and uεkk is a weak solution of −∇ · Ak (x/εk )∇uεkk = 0 on {x ∈ (−1, 1)d−1 , xd ≤ εk ψk (x /εk ) + 1}. For k sufficiently large, by Cacciopoli’s inequality, ˆ

ˆ |∇uεkk |2 dx ≤ C (−1/4,1/4)d

|uεkk |2 dx ≤ C, (−1/2,1/2)d

where C = C(d, N, λ). Therefore, up to a subsequence, which we denote again by uεkk , we have k→∞

uεkk −→ u0 , k→∞

∇uεkk ∇u0 ,

strongly in

L2 ((−1/4, 1/4)d−1 × (−1, 1/4)),

weakly in

L2 ((−1/4, 1/4)d−1 × (−1, 1/4)).

(6.14)

Moreover, εk ψk (·/εk ) converges to 0 because ψk is bounded uniformly in k (see (6.11)). Let ϕ ∈ Cc∞ (D0 (0, 1/4)). Theorem 4 implies that ˆ

k→∞

ˆ

Ak (x/εk )∇uεkk · ∇ϕdx −→ ε

A0 ∇u0 ∇ϕdx, D0 (0,1/4)

Dψk (0,1/4) k

so that u0 is a weak solution to −∇ · A0 ∇u0 = 0 in D0 (0, 1/4). Furthermore, for all ϕ ∈ Cc∞ ((−1/4, 1/4)d−1 × (−1, 0)), ˆ

k→∞

ˆ

uεkk ϕdx −→

0= {x ∈(−1/4,1/4)d−1 , −1≤xd ≤εk ψk (x /εk )}

u0 ϕdx, (−1/4,1/4)d−1 ×(−1,0)

so that u0 (x) = 0 for all x ∈ (−1/4, 1/4)d−1 × (−1, 0). In particular, u0 = 0 in H 1/2 (Δ0 (0, 1/4)). Thus, u0 is a solution to (6.8) and satisfies the estimate (6.9).

C. Kenig, C. Prange / J. Math. Pures Appl. 113 (2018) 1–36

33

εk It remains to pass to the limit in (6.13) to reach a contradiction. Since |Dψ (0, θ)| = |D0 (0, θ)|, we have k

 ⎡     ˆ    ⎢    1   εk εk 0 0  ⎢ ε ∂xd uk − ∂xd u dx (∂xd uk )Dψk (0,θ) − (∂xd u )D0 (0,θ)  ≤ k |D0 (0, θ)| ⎣  Dψεk (0,θ)∩D0 (0,θ)  k ⎤ ˆ   ⎥ ∂x uεk − ∂x u0  dx⎥ . + d k d ⎦ '

ε

Dψk

k

(6.15)

( ' ( ε (0,θ)\D0 (0,θ) ∪ D0 (0,θ)\Dψk (0,θ) k

The first term in the right hand side of (6.15) tends to 0 thanks to the weak convergence of ∇uεkk in (6.14). The second term in the right hand side of (6.15) goes to 0 when k → ∞ because of the L2 bound on the gradient, and the fact that ' ( ' (  k→∞  εk εk (0, θ)  Dψk (0, θ) \ D0 (0, θ) ∪ D0 (0, θ) \ Dψ  −→ 0. k Therefore, ˆ −

 2 " #   k→∞ (∂xd uεkk )Dψεk (0,θ) xd + εk χdk (x/εk ) − (∂xd u0 )D0 (0,θ) xd  −→ 0. k

ε

Dψk (0,θ)∩D0 (0,θ) k

Moreover, the strong L2 convergence in (6.14) implies ˆ −

k→∞

|uεkk − u0 |2 −→ 0.

ε

Dψk (0,θ)∩D0 (0,θ) k

The last thing we have to check is the convergence ˆ k→∞ − |εk vk (x/εk )|2 −→ 0. ε

Dψk (0,θ) k

Let η = η(yd ) ∈ Cc∞ (R) such that η ≡ 1 on (−1, 1) and Supp η ⊂ (−2, 2). We have ˆ −

|εk vk (x/εk )|2

ε

Dψk (0,θ) k

ˆ −

≤

ˆ −

|εk vk (x/εk ) − (xd + εk χdk (x/εk ))η(xd /εk )|2 +

ε

(6.16) |(xd + εk χdk (x/εk ))η(xd /εk )|2 .

ε

Dψk (0,θ)

Dψk (0,θ)

k

k

The last term in the right hand side of (6.16) goes to 0 when k → ∞. Now by Poincaré’s inequality, ˆ −

|εk vk (x/εk ) − (xd + εk χdk (x/εk ))η(xd /εk )|2

ε

Dψk (0,θ) k

⎡

ˆ −

⎢ ≤ Cθ2 ⎢ ⎣ ε

Dψk (0,θ) k

|∇vk (x/εk )|2 +

ˆ −

Dψk (0,θ)

⎤    ⎥ ∇ (xd + εk χdk (x/εk ))η(xd /εk ) 2 ⎥ . ⎦

C. Kenig, C. Prange / J. Math. Pures Appl. 113 (2018) 1–36

34

On the one hand by estimate (6.5) ˆ −

ˆ

Cεd |∇vk (x/εk )| ≤ dk θ

|∇vk (y)|2 dx

2

ε

1 (0,θ/ε ) Dψ k

Dψk (0,θ)

k

k

ˆ

ˆ∞

k→∞

≤ Cεk sup

|∇vk |2 dx ≤ Cεk −→ 0

ξ∈Zd−1

ξ+(0,1)d−1 ψk (y  )

with in the last inequality C = C(d, N, λ, [A]C 0,ν ) uniform in ε, and on the other hand ˆ −

ˆ −

   ∇ (xd + εk χdk (x/εk ))η(xd /εk ) 2 ≤ (1 + ∇χk 2L∞ )

ε

|η(xd /εk )|2

ε

Dψk (0,θ) k

1 + 2 εk

ˆ −

Dψk (0,θ) k

|(xd + εk χdk (x/εk ))η  (xd /εk )|2 ≤ Cεk −→ 0, k→∞

ε

Dψk (0,θ) k

with in the last inequality C = C(d, N, λ, [A]C 0,ν ). These convergence results imply that passing to the limit in (6.13) we get ˆ −

θ2+2μ ≤

 #2 "  εk ε uk (x) − (∂xd ukk )Dψεk (0,θ) xd + εk χdk (x/εk ) + εk vk (x/εk )  dy k

ε

Dψk (0,θ) k

k→∞

−→

ˆ −

 2  0  ) 4, u (x) − (∂xd u0 )D0 (0,θ) xd  ≤ Cθ

D0 (0,θ)

which contradicts (6.10). 6.2. Proof of Lemma 17 The proof is by induction on k. The result for k = 1 is true because of Lemma 16. Let k ∈ N, k ≥ 1. Assume that for all ψ ∈ W 1,∞ (Rd−1 ) such that −1 < ψ < 0 and ∇ψL∞ ≤ γ, for all A ∈ Aν , for all k ∈ N, k > 0, for all 0 < ε < θk−1 ε0 , for all weak solutions uε to (6.2) there exists aεk ∈ RN satisfying |aεk | ≤ C0

1 + θμ + . . . θμ(k−1) , θd/2 (1 − θ)

such that ˆ −

|uε |2 ≤ 1

ε (0,1) Dψ

implies ˆ −

 ε " # u (x) − aεk xd + εχd (x/ε) + εv(x/ε) 2 dy ≤ θ(2+2μ)k .

ε (0,θ k ) Dψ

This is our induction hypothesis.

(6.17)

C. Kenig, C. Prange / J. Math. Pures Appl. 113 (2018) 1–36

35

Given ψ ∈ W 1,∞ (Rd−1 ), −1 < ψ < 0 and ∇ψL∞ ≤ γ and A ∈ Aν , 0 < ε < θk−1 ε0 and a solution uε to (6.2) such that ˆ − |uε |2 ≤ 1 ε (0,1) Dψ

we define U ε (x) :=

#  ε k " 1 u (θ x) − aεk θk xd + εχd (θk x/ε) + εv(θk x/ε) θ(1+μ)k

ε/θ k

for all x ∈ Dψ (0, 1). The goal is to apply the estimate of Lemma 17 to U ε . By the induction estimate (6.17), we have ˆ − |U ε |2 ≤ 1. ε/θ k

Dψ

(0,1)

Moreover, U ε solves the system 

−∇ · A(θk x/ε)∇U ε = 0, ε

U = 0,

ε/θ k

x ∈ Dψ x∈

(0, 1),

ε/θ k Δψ (0, 1).

(6.18)

The boundary layer v solving (6.4) has been designed for U ε to solve (6.18). It follows that U ε satisfies the assumptions of Lemma 16. Therefore, for all ε/θk < ε0 , we have  +2 *  ε ε d k ε k  ≤ θ2+2μ . U (x) − (∂x U ε ) ε/θk + χ (θ x/ε) + v(θ x/ε) x d d   Dψ (0,θ) θk θk

ˆ − ε/θ k

Dψ

(0,θ)

Eventually, ˆ −

 ε " # u (x) − aεk+1 xd + εχd (x/ε) + εv(x/ε) 2 ≤ θ(2+2μ)(k+1) ,

ε (0,θ k+1 ) Dψ

with aεk+1 := aεk + θμk (∂xd U ε )

ε/θ k

Dψ

(0,θ)

satisfying the estimate |aεk+1 | ≤ C0

1 + θμ + . . . θμ(k−1) θμk 1 + θμ + . . . θμk + C0 d/2 ≤ C0 . d/2 θ (1 − θ) θ (1 − θ) θd/2 (1 − θ)

This concludes the iteration step and proves Lemma 17. As a concluding remark, let us notice that Theorem the improved Lipschitz estimate of Theorem 1 can be extended to the system  ε −∇ · A(x/ε)∇uε = f + ∇ · F, x ∈ Dψ (0, 1), (6.19) ε ε x ∈ Δψ (0, 1). u = 0, The proof goes through exactly as in the paper [21].

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36

Theorem 18. Let 0 < μ < 1 and κ > 0. There exists C > 0, such that for all ψ ∈ W 1,∞ (Rd−1 ), for all d2 ×N 2 matrix A = A(y) = (Aαβ , elliptic with constant λ, 1-periodic and Hölder continuous with ij (y)) ∈ R ε ε exponent ν > 0, for all 0 < ε < 1/2, for all f ∈ Ld+κ (Dψ (0, 1)), for all F ∈ C 0,μ (Dψ (0, 1)), for all uε weak solution to (6.19), for all r ∈ [ε, 1/2],

r−d

 εψ(x ˆ /ε)+r

ˆ

|∇uε |2 dxd dx

(−r,r)d−1

≤C

⎧ ⎪ ⎨ ⎪ ⎩

εψ(x /ε)

ˆ

(−1,1)d−1

 εψ(x ˆ /ε)+1

|∇uε |2 dxd dx + f Ld+κ (Dε (0,1)) + F C 0,μ (Dε (0,1))

εψ(x /ε)

⎫ ⎪ ⎬ ⎪ ⎭

.

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