Quantitative Homogenization of Elliptic Operators with Periodic Coefficients

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1 Quantitative Homogenization of Elliptic Operators with Periodic Coefficients Zhongwei Shen Abstract. These lecture notes introduce the quantitative homogenization theory for elliptic partial differential equations. We consider a family of second-order elliptic operators in divergence form with rapidly oscillating periodic coefficients. The qualitative theory of homogenization is presented in Section 1. In Sections 2-5 we address the basic questions in the quantitative theory: convergence rates and uniform regularity estimates. The material in the lecture notes is taken from the monograph [11], where further results and additional topics may be found. 1. Introduction and Qualitative Theory Partial differential equations with rapidly oscillating coefficients are used to model various physical phenomena in heterogeneous media, such as composite and perforated materials. Let ε > 0 be a small parameter, representing the scale of the microstructure of an inhomogeneous medium. The local characteristics of the medium are described by functions of form Ax/ε. Since ε is much smaller than the linear size of the domain where the physical process takes place, solving the corresponding boundary value problems for the partial differential equations directly by numerical methods could be costly. To describe the basic idea of homogenization, we consider the Dirichlet problem, divax/ε uε = F in, u ε = f on, where is a bounded domain in R d. The d d coefficient matrix A = Ay is assumed to be real, bounded measurable, and satisfy the ellipticity condition, Aξ ξ µ ξ 2 for any ξ R d, where µ > 0. Given F H 1 and f H 1/2, the Dirichlet problem has a unique weak solution u ε in H 1. Under the additional structure condition that A is periodic with respect to some lattice in R d, it can be shown that as ε 0, 2010 Mathematics Subject Classification. Primary 35B27; Secondary 35J57, 74Q05. Key words and phrases. Homogenization; Regularity; Convergence Rates. Supported in part by NSF grant DMS copyright holder

2 2 Quantitative Homogenizationof Elliptic Operators with Periodic Coefficients the solution u ε of converges to u 0 weakly in H 1 and thus strongly in L 2. Moreover, the limit u 0 is the weak solution of the Dirichlet problem, div  u 0 = F in, u 0 = f on, where the coefficient matrix  is constant and satisfies the ellipticity condition As a result, one may use the function u 0 as an approximation of u ε. Since  is constant, the Dirichlet problem is much easier to handle both analytically and numerically than Similar results may be proved for Neumann type boundary value problems. The entries of the constant matrix Â, which are uniquely determined by A, are given by solving some auxiliary problems in a periodic cell. By taking the inhomogeneity scale ε to zero in limit, we have "homogenized" the microscopically inhomogeneous medium. The limit u 0 of the solution u ε is called the homogenized or effective solution. The boundary value problem is referred as the homogenized problem for The constant matrix Â, which describes the macroscopic characteristics of the inhomogeneous medium, is called the matrix of homogenized or effective coefficients. See [3, 15] for homogenization of partial differential equations in various settings. In this section we establish the qualitative homogenization theory for the elliptic operator divax/ε. Section 2 is devoted to the question of convergence rates. In Sections 3, 4 and 5, we investigate various regularity estimates that are uniform with respect to the small parameter ε. Throughout the notes we will use C and c to denote positive constants that are independent of the parameter ε > 0. They may change from line to line and depend on A and/or. We will use ffl E u to denote the L1 average of a function u over a set E; i.e. u = 1 u. E E E The summation convention that the repeated indices are summed will be used throughout Correctors and effective coefficients Let L ε = div A x/ε, ε > 0 in R d. The coefficient matrix tensor A in is given by Ay = a αβ ij y, with 1 i, j d and 1 α, β m. We will alway assume that A is real, bounded measurable, and satisfies the conditions, µ ξ 2 a αβ ij ξα i ξβ j and a αβ ij µ 1 for any ξ = ξ α i Rm d, where µ > 0. For simplicity, from now on, we shall suppress all super indices or assume m = 1; however, results and proofs

3 Zhongwei Shen 3 extend to the case m > 1 - the case of elliptic systems. We also assume that A is 1-periodic; i.e., for each z Z d, Ay + z = Ay for a.e. y R d. By a linear transformation one may replace Z d in by any lattice in R d. For Y = [0, 1 d and k 1, let H k pery denote the closure in H k Y of 1-periodic C functions in R d. Define a per φ, ψ = Y A φ ψ dy for φ, ψ H 1 Y. By the Lax-Milgram Theorem, for each 1 j d, there exists a unique function χ j in H 1 pery such that χ j dy = 0 and a per χj, ψ = a per yj, ψ for any ψ H 1 pery. Y Note that χ j H 1 pery is the unique weak solution of the following problem: Lχ j = Ly j in R d, χ j is 1-periodic, χ j dy = 0, where L = divay. In particular, if Y then χ j = 0 in R d. Ly j = a ij y i = 0, Definition The vector-valued function χ = χ 1, χ 2,..., χ d H 1 Y; R d is called the first-order corrector for the operator L ε. It follows from with ψ = χ j that χ j H 1 Y C, where C depends only on d and µ. Since Ly j + χ j = 0 in R d, by rescaling, we obtain L ε xj + εχ j x/ε = 0 in R d for any ε > 0. Definition Let  = â ij, where 1 i, j d and χ j â ij = a ij + a ik dy, Y y k and define L 0 = divâ.

4 4 Quantitative Homogenizationof Elliptic Operators with Periodic Coefficients The constant matrix  is called the matrix of effective or homogenized coefficients. The operator L 0 is called the homogenized operator. The following theorem gives the ellipticity for the homogenized operator L 0. Theorem Suppose that A is 1-periodic and satisfies Then µ ξ 2 Âξ ξ and  µ 1 for any ξ R d, where µ 1 > 0 depends only on d and µ. Proof. The second inequality in follows readily from To see the first, we fix ξ = ξ i R d and let φ = ξ i y i, ψ = ξ i χ i. Observe that â ij = a per y j + χ j, y i = a per y j + χ j, y i + χ i. By 1.1.2, â ij ξ i ξ j = a per φ + ψ, φ + ψ µ = µ Y Y φ + ψ 2 dy φ 2 dy + µ Y ψ 2 dy, where we have used the fact Y χ i dy = 0. It follows that â ij ξ i ξ j µ Y = µ ξ 2. φ 2 dy Theorem Let A = a ij denote the adjoint of A, where a ij = a ji. Then  = Â. In particular, if A is symmetric, i.e. a ij y = a ji y for 1 i, j d, so is Â. Proof. Let χ y = χ j y denote the corrector for L ε; i.e. function in H 1 pery such that Y χ j dy = 0 and χ j a perχ j, ψ = a pery j, ψ for any ψ H 1 pery, where a perφ, ψ = a per ψ, φ. Observe that by and , â ij = a per yj + χ j, y i = aper yj + χ j, y i + χ i = a per yi + χ i , y j + χ j = a per yi + χ i, y j = a per yi + χ i, y j + χ j = â ji, for 1 i, j d. is the unique

5 Zhongwei Shen H-Compactness Proposition Let h l be a sequence of 1-periodic functions. Assume that Let ε l 0. Then h l L 2 Y C and Y h l dy c 0 as l. h l x/ε l c 0 weakly in L 2 as l, where is a bounded domain in R d. In particular, if h is 1-periodic and h L 2 Y, then hx/ε Y h dy weakly in L 2, as ε 0. Proof. By considering the periodic functions h l ffl Y h l, one may assume that Y h l = 0 and hence c 0 = 0. Let u l H 2 pery be a 1-periodic function such that Let g l = u l. Then h l = divg l and Note that u l = h l in Y. g l L 2 Y C h l L 2 Y C. h l x/ε l = ε l div g l x/ε l. It follows that if ϕ C 1 0, h l x/ε l ϕx dx = ε l g l x/ε l ϕx dx 0, as ε l 0. To see this, we fix R > 0 so that B0, R. Then g l x/ε l 2 dx ε d l g l y 2 dy B0,R/ε l C g l 2 L 2 C, Y where we have used the periodicity of g l for the second inequality the constant C depends on R. Similarly, h l x/ε l L 2 C h l L 2 Y C; i.e., the sequence h l x/ε l is bounded in L 2. In view of we conclude that h l x/ε l 0 weakly in L 2. Recall that for u = u 1, u 2,..., u d, divu = d i=1 u i x i and curlu = u i uj. x j x i 1 i,j d The following theorem is usually referred as the Div-Curl Lemma. Theorem Let u l and v l be two bounded sequences in L 2 ; R d. Suppose that 1 u l u and v l v weakly in L 2 ; R d ; 2 curlu l = 0 in and divv l f strongly in H 1.

6 6 Quantitative Homogenizationof Elliptic Operators with Periodic Coefficients Then u l v l ϕ dx as l, for any scalar function ϕ C 1 0. u v ϕ dx, The next theorem shows that the sequence of operators L l ε l is H-compact in the sense of H-convergence [14]. Theorem Let A l y be a sequence of 1-periodic matrices satisfying with the same constant µ. Let F l H 1. Suppose that L l ε l u l = F l in, where ε l 0, u l H 1, and L l ε l = div A l x/ε l. We further assume that F l F in H 1, u l u weakly in H 1, Â l A 0, where Âl denotes the matrix of effective coefficients for A l. Then A l x/ε l u l A 0 u weakly in L 2 ; R d, A 0 is a constant matrix satisfying the ellipticity condition , and u is a weak solution of diva 0 u = F in. Proof. Since Âl A 0 and Âl satisfies , so does A 0. Also, follows directly from and To see 1.2.5, we let u l be a subsequence such that A l x/ε l u l H weakly in L 2 ; R d for some H L 2 ; R d and show that H = A 0 u. This would imply that the full sequence A l x/ε l u l converges weakly to A 0 u in L 2 ; R d. Without loss of generality we assume that A l x/ε l u l H weakly in L 2 ; R d for some H = H i L 2 ; R d. Let χ l y = χ k,l y denote the correctors associated with the matrix A l, the adjoint of A l. Fix 1 k d and consider the identity A l x/ε l u l x k + ε l χ k,l x/ε l ψ dx = u l A l x/ε l x k + ε l χ k,l x/ε l ψ dx,

7 Zhongwei Shen 7 where ψ C 1 0. By Proposition 1.2.1, x k + ε l χ k,l x/ε l = x k + χ k,l x/ε l x k weakly in L 2, where we have used the fact Y χ k,l dy = 0. Since Ll ε l u εl = F l in, in view of and 1.2.9, it follows by Theorem that the left-hand side LHS of converges to H x k ψ dx = H k ψ dx. Similarly, note that u l u weakly and A l x/ε l x k + ε l χ k,l x/ε l lim l Y A l = lim l  l x k =A 0 x k x k + χ k,l dy weakly in L 2, where we have used Proposition as well as Theorem Since L l ε l xk + ε l χ k,l x/ε l = 0 in R d, we may use Theorem again to claim that the RHS of converges to u A 0 x k ψ dx. As a result, since ψ C 1 0 is arbitrary, it follows that H k = u A 0 x k = A 0 u x k in. This shows that H = A 0 u and completes the proof. We now use Theorem to establish the qualitative homogenization of the Dirichlet and Neumann problems for L ε. The proof only uses a special case of Theorem 1.2.3, where A l = A is fixed. The general case is essential in a compactness argument we will use in Section 3 for the large-scale regularity Homogenization of boundary value problems Assume that A satisfies and is 1-periodic. Let F L 2, G L 2 ; R d and f H 1/2. There exists a unique u ε H 1 such that L ε u ε = F + divg in and u ε = f on the boundary data is taken in the sense of trace. Furthermore, the solution u ε satisfies u ε H 1 F C L 2 + G L 2 + f H 1/2, where C depends only on µ and. Let u ε be a subsequence of u ε such that as ε 0, u ε u weakly in H 1

8 8 Quantitative Homogenizationof Elliptic Operators with Periodic Coefficients for some u H 1. It follows by Theorem that Ax/ε u ε Â u and L 0 u = F + divg in. Since f H 1/2, there exists Φ H 1 such that Φ = f on. Using the facts that u ε Φ u Φ weakly in H 1 and u ε Φ H 1 0, we see that u Φ H1 0. Hence, u = f on. Consequently, u is the unique weak solution to the Dirichlet problem, L 0 u 0 = F + divg in and u 0 = f on. Since u ε is bounded in H 1 and thus any sequence u εl with ε l 0 contains a subsequence that converges weakly in H 1, one may conclude that as ε 0, Ax/ε uε Â u 0 weakly in L 2 ; R d, u ε u 0 weakly in H 1. By the compactness of the embedding H 1 L 2, we also obtain u ε u 0 strongly in L 2. Let F L 2, G L 2 ; R d, and g H 1/2, the dual of H 1/2. Consider the Neumann problem L ε u ε = F + divg in, u ε = g n G on, ν ε where the conormal derivative u ε ν ε on is defined by u ε = n i xa ij x/ε u ε, ν ε x j and n denotes the outward unit normal to. Definition We call u ε H 1 a weak solution of the Neumann problem if Ax/ε u ε ϕ dx = Fϕ dx G ϕ dx + g, ϕ H 1/2 H 1/2 for any ϕ C R d. Let u ε be a subsequence of u ε such that u ε u 0 weakly in H 1 for some u 0 H 1. It follows by Theorem that Ax/ε u ε Â u 0 weakly in L 2 ; R d.

9 Zhongwei Shen 9 By taking limits in we see that u 0 is a weak solution to the Neumann problem: u L 0 u 0 = F + divg in and = g n G on, ν 0 and that u 0 dx = 0, where u 0 ν 0 = n i â ij u 0 x j is the conormal derivative associated with the operator L 0. Since such u 0 is unique, we conclude that as ε 0, u ε u 0 weakly in H 1 and thus strongly in L 2. As in the case of the Dirichlet problem, we also obtain Ax/ε u ε  u 0 weakly in L 2 ; R d. Problem Find explicit formulas for χ and  in the case d = 1. Problem Prove Theorem Hint: If curlg = 0 in a ball B, then there exists G H 1 B such that g = G in B. Problem Assume that F, G and g satisfy the compatibility condition F dx + g, 1 H 1/2 H 1/2 = 0. Show that the Neumann problem has a unique up to a constant in R solution. Furthermore, if u ε dx = 0, then u ε H 1 F C L 2 + G L 2 + g H 1/2, where C depends only on µ and. 2. Convergence Rates Let L ε = divax/ε for ε > 0, where Ay = a ij y is 1-periodic and satisfies the ellipticity condition Let L 0 = divâ, where  = â ij denotes the matrix of effective coefficients, given by For F L 2 and ε 0, consider the Dirichlet problem Lε u ε = F in, u ε = f on, and the Neumann problem L ε u ε = F in, u ε = g on, ν ε u ε R in L 2, where f H 1, and g L 2. It is shown in Section 1 that as ε 0, u ε converges to u 0 weakly in H 1 and strongly in L 2. In this section we investigate the problem of convergence rates in H 1 and L p.

10 10 Quantitative Homogenizationof Elliptic Operators with Periodic Coefficients 2.1. Flux correctors and ε-smoothing For 1 i, j d, let b ij y = a ij y + a ik y χj y â ij, y k where the repeated index k is summed from 1 to d. Observe that the matrix By = b ij y is 1-periodic and that B L p Y C 0 for some p > 2 and C 0 > 0 depending on d and µ. Moreover, it follows from the definitions of χ j and â ij that y i bij = 0 and Y b ij y dy = 0. Proposition There exist φ kij H 1 pery, where 1 i, j, k d, such that b ij = φkij and φ kij = φ ikj in Y. y k Moreover, if χ = χ j is Hölder continuous, then φ kij L Y. Proof. Since Y b ij dy = 0, there exists f ij H 2 pery such that Y f ij dy = 0 and f ij = b ij in Y. Moreover, Define f ij H 2 Y C b ij L 2 Y φ kij y = C. fij fkj. y k y i Clearly, φ kij H 1 pery and φ kij = φ ikj. Using y bij i = 0, we deduce from that yi fij is a 1-periodic harmonic function and thus is constant. Hence, 2 φkij = fij fkj y k y k y i = f ij = b ij. Suppose that the corrector χ is Hölder continuous. Recall that By Caccioppoli s inequality, χ 2 dx C r 2 By,r L χ j + y j = 0 in R d. By,2r χx χy 2 dx + C r d. This implies that χ is in the Morrey space L 2,ρ Y for some ρ > d 2; i.e., χ 2 dx C r ρ for y Y and 0 < r < 1. By,r Consequently, b ij L 2,ρ Y for some ρ > d 2 and b ij y dy C. x y d 1 sup x Y Y

11 Zhongwei Shen 11 In view of 2.1.4, by considering f ij ϕ, where ϕ is a cut-off function, and using a potential representation for Laplace s equation, one deduces that f ij L Y C f ij L 2 Y + C sup b ij y dy x Y Y x y d 1 C. It follows that φ kij L Y. Remark Fro scalar elliptic equations in divergence form, it follows from the classical De Giorgi - Nash estimate that χ is Hölder continuous. As a result, the flux corrector is bounded. Remark A key property of φ = φ kij, which follows from 2.1.3, is the identity: b ij x/ε ψ = ε φ kij x/ε ψ x i x k x i for any ψ H 2. Let u ε H 1, u 0 H 2, and A direct computation shows that w ε = u ε u 0 ε χx/ε u 0. Ax/ε u ε Â u 0 Bx/ε u 0 = Ax/ε w ε + ε Ax/εχx/ε 2 u 0, where By = b ij y is defined by It follows that Ax/ε u ε Â u 0 Bx/ε u 0 L 2 C w ε L 2 + C ε χx/ε 2 u 0 L 2 C u ε u 0 χx/ε u 0 L 2 + Cε χx/ε 2 u 0 L 2, where C depends only on µ. This indicates that the 1-periodic matrix-valued function By plays the same role for the flux Ax/ε u ε as χy does for u ε. Since b ij = y φkij k, the 1-periodic function φ = φkij is called the flux corrector. To deal with the fact that the correctors χ and φ may be unbounded for elliptic systems in higher dimensions, we introduce an ε-smoothing operator S ε. Definition Fix ρ C 0 B0, 1/2 such that ρ 0 and R d ρ dx = 1. For ε > 0, define S ε fx = ρ ε fx = fx yρ ε y dy, R d where ρ ε y = ε d ρy/ε.

12 12 Quantitative Homogenizationof Elliptic Operators with Periodic Coefficients Proposition Let f L p loc Rd for some 1 p <. L p loc Rd, gx/ε S ε f L p O C sup x R d where O R d is open, and C depends only on p. Ot = x R d : distx, O < t, Proof. By Hölder s inequality, S ε fx p fy p ρ ε x y dy, R d where we also used the fact ρ ε dx = 1. R d Then for any g 1/p g p f L p Oε/2, Bx,1/2 This, together with Fubini s Theorem, gives for the case O = R d. The general case follows from the observation that for any x O. S ε fx = S ε fχ Oε/2 x It follows from that if g is 1-periodic and belongs to L p Y, then gx/εs ε f L p O C g L p Y f L p Oε/2, where C depends only on d and p. Proposition Let f W 1,p R d for some 1 p. Then S ε f f L p R d ε f L p R d. Moreover, if q = 2d d+1, where C depends only on d. Proof. Using S ε f L 2 R d Cε 1/2 f L q R d, S ε f f L 2 R d Cε1/2 f L q R d, fx + y fx = 1 0 fx + ty y dt and Minkowski s inequality for integrals, we see that f + y f L p R d y f L p R d

13 for any y R d. By Minkowski s inequality again, Zhongwei Shen 13 S ε f f L p R d R f εy f L p d R d ρy dy ε y ρy dy f L p R d R d ε f L p R d, which gives Next, note that the Fourier transform of S ε f is given by ρεξ fξ. Let q = 2d d+1. By Plancheral s theorem and Hölder s inequality, S ε f 2 dx = ρεξ 2 fξ 2 dξ R d R d 1/d ρεξ 2d dξ f 2 R d L q R d Cε 1 f 2 L q R d, where we have used the Hausdorff-Young inequality f L q R d f L q R d in the last step. This gives the first inequality in Similarly, using ρ0 = ρ dx = 1, R d we obtain S ε f f L 2 R d = ρεξ 1 f L 2 R d d C ρεξ ρ0 2d ξ 2d dξ f R d Lq R d = Cε 1/2 ρξ ρ0 ξ 1 L 2d R d f L q R d Cε 1/2 f L q R d, where we have also used the fact ρξ ρ0 C ξ for ξ 1 in the last step. Let t = x : distx, < t, where t > 0. Proposition Let be a bounded Lipschitz domain in R d and q = Then for any f W 1,q, f L 2 t Ct1/2 f W 1,q and f L 2 C f W 1,q, where C depends only on. 2d d+1.

14 14 Quantitative Homogenizationof Elliptic Operators with Periodic Coefficients Proposition Let be a bounded Lipschitz domain in R d and q = d+1 2d. Let g L 2 loc Rd be a 1-periodic function. Then for any f W 1,q, gx/ε 2 S ε f 2 dx Ct g 2 L 2t \ 2 Y f 2 W 1,q, t where t ε and C depends only on. We end this section with the fractional integral estimates. Proposition For 0 < α < d, define T α fx = Then, if 1 < p < d α and 1 q = 1 p α d, R d fy dy. x y d α T α f L q R d C f L p R d, where C depends only on α and p. Proof. The inequality is referred as the Hardy-Littlewood-Sobolev inequality. We refer the reader to [7, pp ] for its proof Convergence rates in H 1 Fix a cut-off function η ε C 0 0 η ε 1, η ε C/ε, η ε x = 1 if distx, 4ε, η ε x = 0 if distx, 3ε. such that Let S 2 ε = S ε S ε and define w ε = u ε u 0 εχx/εη ε S 2 ε u 0, where u ε H 1 is the weak solution of Dirichlet problem and u 0 the homogenized solution. Lemma Let be a bounded Lipschitz domain in R d and t be defined by Then, for any ψ H 1 0, Ax/ε w ε ψ dx C ψ L 2 ε S ε 2 u 0 L 2 \ 3ε + u 0 S ε u 0 L 2 \ 2ε + C ψ L 2 4ε u 0 L 2 5ε, where w ε is given by and C depends only on µ and. Proof. Since w ε H 1, by a density argument, it suffices to prove for any ψ C 0. Note that Ax/ε w ε = Ax/ε u ε Ax/ε u 0 Ax/ε χx/εη ε S 2 ε u 0 εax/εχx/ε η ε S 2 ε u 0 = Ax/ε u ε Â u 0 + Â Ax/ε [ u 0 η ε S 2 ε u 0 ] Bx/εη ε S 2 ε u 0 εax/εχx/ε η ε S 2 ε u 0,

15 Zhongwei Shen 15 where we have used the fact By = Ay + Ay χy Â. Using Ax/ε u ε ψ dx = Â u 0 ψ dx for any ψ C 0, we obtain Ax/ε w ε ψ dx C 1 η ε u 0 ψ dx + C η ε u 0 S 2 ε u 0 ψ dx η ε Bx/εS 2 ε u 0 ψ dx + Cε χx/ε η ε S 2 ε u 0 ψ dx. Since η ε = 1 in \ 4ε, by the Cauchy inequality, the first term in the RHS of is bounded by C u 0 L 2 4ε ψ L 2 4ε. Using η ε = 0 in 3ε and u 0 S 2 ε u 0 L 2 \ 3ε u 0 S ε u 0 L 2 \ 3ε + S ε u 0 S 2 ε u 0 L 2 \ 3ε C u 0 S ε u 0 L 2 \ 2ε, we may bound the second term by C u 0 S ε u 0 L 2 \ 2ε ψ L 2. Also, by the Cauchy inequality and , the fourth term in the RHS of is dominated by C u 0 L 2 5ε ψ L 2 4ε + Cε S ε 2 u 0 L 2 \ 2ε ψ L 2. Finally, to handle the third term in the RHS of 2.2.6, we use the identity to obtain η ε Bx/εS 2 ε u 0 ψ = b ij x/εs 2 u0 ψ ε η ε x j x i = ε φ kij x/ε ψ S 2 u0 ε η ε. x k x i x j

16 16 Quantitative Homogenizationof Elliptic Operators with Periodic Coefficients It follows from and integration by parts that η ε Bx/εS 2 ε u 0 ψ dx Cε η ε φx/ε ψ S 2 ε 2 u 0 dx + Cε η ε φx/ε ψ S 2 ε u 0 dx Cε ψ L 2 S ε 2 u 0 L 2 \ 2ε + C ψ L 2 4ε u 0 L 2 5ε, where we have used the Cauchy inequality and for the last step. Theorem Assume that A is 1-periodic and satisfies Let be a bounded Lipschitz domain in R d. Let w ε be given by Then for 0 < ε < 1, w ε L 2 ε C 2 u 0 L 2 \ ε + u 0 L 2 5ε. Consequently, w ε H 1 0 C ε u 0 H 2. The constant C depends only on µ and. Proof. Since w ε H 1 0, we may take ψ = w ε in This, together with the ellipticity condition 1.1.2, gives w ε L 2 Cε 2 u 0 L 2 \ 2ε + C u 0 S ε u 0 L 2 \ 2ε + C u 0 L 2 5ε. Choose η ε C 0 such that 0 η ε 1, η ε = 0 in ε, η ε = 1 in \ 3ε/2, and η Cε 1. It follows that u 0 S ε u 0 L 2 \ 2ε η ε u 0 S ε η ε u 0 L 2 R d Cε η ε u 0 L 2 R d C ε 2 u 0 L 2 \ ε + u 0 L 2 2ε, where we have used for the second inequality. The estimate now follows from and Note that, by , u 0 L 2 5ε C ε u 0 H 2. The inequality follows from and Under the additional symmetry condition A = A, i.e., a ij = a ji, an improved estimate can be obtained using sharp regularity estimates for L 0. Theorem Suppose that A is 1-periodic and satisfies Also assume that A = A. Let be a bounded Lipschitz domain in R d. Let w ε be given by Then, for 0 < ε < 1, w ε H 1 0 C ε F L q + f H 1, where q = d+1 2d and C depends only on µ and.

17 Zhongwei Shen 17 Definition For a continuous function u in a bounded Lipschitz domain, the nontangential maximal function of u is defined by u x = sup uy : y and y x < C 0 disty, for x, where C 0 > 1 is a sufficiently large constant depending on. The proof of Theorem relies on the following regularity result for solutions of L 0 u = 0 in. Lemma Assume that A satisfies the same conditions as in Theorem Let be a bounded Lipschitz domain in R d. Let u H 1 be a weak solution to the Dirichlet problem: L 0 u = 0 in and u = f on, where f H 1. Then u L 2 C f H 1, where C depends only on µ and. Proof. By Lemmas and 1.1.4, Â satisfies the ellipticity condition and is symmetric. As a result, the estimate follows from [5, 6, 8]. Proof of Theorem We start by taking ψ = w ε H 1 0 in By the ellipticity condition 1.1.2, this gives w ε L 2 Cε S ε 2 u 0 L 2 \ 3ε + C u 0 L 2 5ε + C u 0 S ε u 0 L 2 \ 2ε. To bound the RHS of , we write u 0 = v 0 + φ, where v 0 x = Γ 0 x yfy dy and Γ 0 x denotes the fundamental solution for the homogenized operator L 0 in R d, with pole at the origin. It follows from the standard Calderón-Zygmund estimates for singular integrals see [12, Chapter 2] that v 0 L r R d C F L r, for any 1 < r <. Since Γ 0 x C x 1 d, by the fractional integral estimates in Proposition 2.1.9, v 0 L p R d C F L q, 2d where p = d 1 and q = p = , yield that d+1 2d. These estimates, together with and and that ε S ε 2 v 0 L 2 \ 3ε Cε1/2 2 v 0 L q R d Cε 1/2 F L q, v 0 L 2 5ε Cε1/2 v 0 W 1,q Cε 1/2 F L q.

18 18 Quantitative Homogenizationof Elliptic Operators with Periodic Coefficients Also, note that by , v 0 S ε v 0 L 2 \ 2ε Cε1/2 2 v 0 L q R d In summary we have proved that Cε 1/2 F L q. ε S ε 2 v 0 L 2 \ 3ε + v 0 L 2 5ε + v 0 S ε v 0 L 2 \ 2ε Cε1/2 F L q. It remains to bound the LHS of , with v 0 replaced by φ. To this end we first note that L 0 φ = 0 in and φ = f v 0 on. This allows us to apply Lemma Since v 0 H 1 C v 0 W 2,q C F L q, where we have used for the first inequality, we obtain It follows that φ L 2 f C H 1 + v 0 H 1 C f H 1 + F L q. φ L 2 5ε Cε1/2 φ L 2 Cε 1/2 f H 1 + F L q. Next, we use the interior estimate for L 0, 2 φx 2 C r d+2 Bx,r φy 2 dy, where r = distx, /8, and Fubini s Theorem to obtain 2 φx 2 dx C φx 2[ distx, ] 2 dx \ ε \ ε/2 Cε 1 φ 2 dσ. Hence, ε S ε 2 φ L 2 \ 3ε Cε 2 φ L 2 \ 2ε Finally, we observe that as in , Cε 1/2 φ L 2 Cε 1/2 f H 1 + F L q. φ S ε φ L 2 \ 2ε ε C 2 φ L 2 \ ε + φ L 2 2ε Cε 1/2 f H 1 + F L q,

19 Zhongwei Shen 19 where we have used for the second inequality. As a result, we have proved that ε S ε 2 φ L 2 \ 3ε + φ L 2 5ε + φ S ε φ L 2 \ 2ε Cε 1/2 f H 1 + F L q. This, together with and , gives Remark The proof of Theorem only uses the symmetry of Â. Thus we may drop the assumption A = A in the case of scalar elliptic equations, as L 0 u = 1/2L 0 + L 0 u. Also, since F L q = L 0 u 0 L q C 2 u 0 L q and f H 1 C u 0 W 2,q, it follows from that w ε H 1 0 C ε u 0 W 2,q, where C depends only on µ and. We now consider the two-scale expansions without the ε-smoothing. Theorem Assume that A is 1-periodic and satisfies Also assume that χ is bounded. Let be a bounded Lipschitz domain in R d. Let u ε H 1 be the weak solution of Dirichlet problem and u 0 the homogenized solution. Then, u ε u 0 εχx/ε u 0 H 1 C ε u 0 H 2, where 0 < ε < 1 and C depends only on µ, χ and. Lemma If χ is bounded, then χx/εψ L 2 C1 + χ ε ψ L 2 + ψ L 2 for any ψ H 1, where C depends only on µ and. Proof. Let u ε = εχ j x/ε + x j x 0 j. Since L ε u ε = 0 in R d, it follows by Caccioppoli s inequality that u ε 2 ϕ 2 dx C u ε 2 ϕ 2 dx R d R d for any ϕ C 1 0 Rd. Thus, if ϕ C 1 0 Bx 0, 2ε, χx/ε 2 ϕ 2 dx C ϕ 2 + Cε 2 Bx 0,2ε Bx 0,2ε + Cε 2 Bx 0,2ε Bx 0,2ε ϕ 2 dx χx/ε 2 ϕ 2 dx.

20 20 Quantitative Homogenizationof Elliptic Operators with Periodic Coefficients Let ϕ = ψ η ε, where ψ C 1 0 Rd and η ε is a cut-off function in C 1 0 Bx 0, 2ε with the properties that 0 η ε 1, η ε = 1 on Bx 0, ε and η ε C/ε. We obtain χx/ε 2 ψ 2 dx C 1 + χx/ε 2 ψ 2 dx Bx 0,ε Bx 0,2ε + Cε 2 Bx 0,2ε 1 + χx/ε 2 ψ 2 dx. By integrating the inequality above in x 0 over R d we see that χx/ε 2 ψ 2 dx R d C 1 + χx/ε 2 ψ 2 dx + Cε χx/ε 2 ψ 2 dx R d R d for any ψ C 1 0 Rd. If χ is bounded, it follows from that χx/εψ L 2 C1 + χ ψ L 2 R d + ε ψ L 2 R d for any ψ C 1 0 Rd. By a limiting argument, the inequality holds for any ψ H 1 R d. This gives , using the fact that for any ψ H 1, one may extend it to a function ψ in H 1 R d so that ψ L 2 R d C ψ L 2 and ψ H 1 R d C ψ H 1 [12, Chpater VI]. Proof of Theorem Suppose χ is bounded , it suffices to show that To prove , in view of εχx/ε u 0 εχx/εη ε S 2 ε u 0 H 1 C ε u 0 H 2. To this end, we note that the LHS of is bounded by Cε χx/ε u 0 η ε S 2 ε u 0 L 2 + C χx/ε u 0 η ε S 2 ε u 0 L 2 + Cε χx/ε u 0 η ε S 2 ε u 0 L 2 Cε u 0 η ε S 2 ε u 0 L 2 + C u 0 η ε S 2 ε u 0 L 2, where we have used Note that ε u 0 η ε S 2 ε u 0 L 2 ε 2 u 0 L 2 + ε η ε S 2 ε u 0 L 2 and Cε u 0 H 2 + C u 0 L 2 5ε C ε u 0 H 2, u 0 η ε S 2 ε u 0 L 2 C u 0 L 2 5ε + u 0 S 2 ε u 0 L 2 \ 4ε C ε u 0 H 2, where we also used for the last inequality. Theorem Assume that A is 1-periodic and satisfies Let be a bounded Lipschitz domain in R d. Let u ε H 1 be the weak solution of the

21 Zhongwei Shen 21 Neumann problem with u ε dx = 0. Let u 0 be the homogenized solution. Then, if the corrector χ is bounded and u 0 H 2, u ε u 0 εχx/ε u 0 H 1 C ε u 0 H 2 for any 0 < ε < 1, where C depends only on µ, χ and Convergence rates in L p In this subsection we establish a sharp Oε convergence rate in L p with p = d 1 2d for Dirichlet problem 2.0.1, under the assumptions that A is 1-periodic, satisfies and A = A. Without the symmetry condition we obtain an Oε convergence rate in L 2. Lemma Let be a bounded Lipschitz domain in R d. Let w ε be given by 2.2.2, where u ε is the weak solution of and u 0 the homogenized solution. Assume that u 0 W 2,q for q = d+1 2d. Then, for any ψ C 0, Ax/ε w ε ψ dx R d C u 0 W 2,q ε ψ L p + ε ψ L 2 4ε, where p = q = d 1 2d and C depends only on µ and. Proof. An inspection of the proof of Lemma shows that Ax/ε w ε ψ dx C ψ L p ε S ε 2 u 0 L q \ 3ε + u 0 S ε u 0 L q \ 2ε + C ψ L 2 4ε u 0 L 2 5ε. Note that u 0 L 2 5ε C ε u 0 W 2,q and S ε 2 u 0 L q \ 3ε C 2 u 0 L q. Since u 0 W 2,q, there exists ũ 0 W 2,q R d such that ũ 0 = u 0 in and ũ 0 W 2,q R d C u 0 W 2,q. It follows that u 0 S ε u 0 L q \ 3ε ũ 0 S ε ũ 0 L q R d Cε 2 ũ 0 L q R d Cε u 0 W 2,q, where we have used for the second inequality. In view of we have proved Lemma Assume that  = Â. Let be a bounded Lipschitz domain in R d. Let u H 1 0 be a weak solution to the Dirichlet problem: L 0u = F in and u = 0 on, where F C 0. Then u L 2 t Ct1/2 F L q,

22 22 Quantitative Homogenizationof Elliptic Operators with Periodic Coefficients for 0 < t < diam, and u L p C F L q, where p = d 1 2d, q = p = d+1 2d, and C depends only on µ and. Proof. Write u = φ + v 0, where v 0 is defined by The proof of is essentially contained in that of Theorem We leave it as an exercise for the reader. To see 2.3.5, we recall that by the fractional integral estimates, v 0 L p C F L q. To estimate φ, we use the following inequality 1/p 1/ ψ dx p C ψ dσ 2, where p = 2d d 1, ψ is a continuous function in and ψ denotes the nontangential maximal function of ψ. This gives φ L p C φ L 2 C v 0 L 2 C v 0 W 2,q C F L q, where we have used Lemma for the second inequality and for the third. Finally, to prove 2.3.6, we observe that for any x and x with x x = distx, = r, ψx ψ y, if y and y x cr. It follows that ψx C r d 1 C Let f C 1 0 and gy = Note that by 2.3.7, ψxfx dx C = C 2d d+1 B x,cr ψ dσ ψ y dσy. x y d 1 fx dx. x y d 1 ψ y fx x y d 1 dσydx ψ g dσ C ψ L 2 g L 2 C ψ L 2 g W 1,q C ψ L 2 f L q, where q = and we have used for the third inequality and singular integral estimates for the fourth. The inequality follows by a duality argument.

23 Zhongwei Shen 23 The next theorem gives a sharp Oε convergence rate in L p with p = 2d d 1. Theorem Assume that A is 1-periodic and satisfies Also assume that A = A. Let be a bounded Lipschitz domain in R d. Let u ε ε 0 be the weak solution of Assume that u 0 W 2,q for q = 2d d+1. Then u ε u 0 L p Cε u 0 W 2,q, where p = q = d 1 2d and C depends only on µ and. Proof. Let w ε be given by For any G C 0, let v ε H 1 0 ε 0 be the weak solution to the Dirichlet problem, L εv ε = G in and v ε = 0 on. Define r ε = v ε v 0 εχx/εη ε S 2 ε v 0, where η ε is a cut-off function satisfying Observe that w ε G dx = Ax/ε w ε v ε dx Ax/ε w ε r ε dx + Ax/ε w ε v 0 dx + Ax/ε w ε εχx/εη ε S 2 ε v 0 dx = I 1 + I 2 + I 3. To estimate I 1, we note that by and , w ε L 2 C ε u 0 W 2,q and r ε L 2 C ε G L q, where we have used the assumption A = A. By the Cauchy inequality this gives I 1 Cε u 0 W 2,q G L q. Next, to bound I 2, we use Lemma to obtain I 2 C u W 2,q ε v 0 L p + ε v 0 L 2 4ε Cε u 0 W 2,q G L q, where we use Lemma for the last step. To estimate I 3, we let Using as well as the observation it is not hard to show that ϕ ε = εχx/εη ε S 2 ε v 0. S 2 εf L p R d Cε 1 f L p R d, ε ϕ ε L p + ε ϕ ε L 2 4ε Cε v 0 L p + C ε v 0 L 2 5ε.

24 24 Quantitative Homogenizationof Elliptic Operators with Periodic Coefficients As in the case of I 2, by Lemma 2.3.1, this implies that I 3 C u W 2,q ε ϕ ε L p + ε ϕ ε L 2 4ε Cε u 0 W 2,q G L q. In view of we have proved that w ε G dx Cε u 0 W 2,q G L q, where C depends only on µ and. By duality this implies that w ε L p Cε u 0 W 2,q. It follows that u ε u 0 L p w ε L p + εχx/εη ε S 2 ε u 0 L p Cε u 0 W 2,q. The next theorem gives the Oε convergence rate in L 2 without the symmetry condition, assuming that is a bounded C 1,1 domain The smoothness assumption on ensures the H 2 estimate for L 0. Theorem Assume that A is 1-periodic and satisfies Let be a bounded C 1,1 domain in R d. Let u ε H 1 ε 0 be the weak solution of Assume that u 0 H 2. Then u ε u 0 L 2 Cε u 0 H 2, where C depends only on µ and. Proof. The proof is similar to that of Theorem By Theorem 2.2.2, the estimates w ε H 1 0 C ε u 0 H 2 and r ε H 1 0 C ε v 0 H 2 hold without the symmetry condition. It follows from the proof of Theorem that w ε G dx Cε u 0 H 2 v 0 H 2. Recall that v 0 H 1 0 is a solution of L 0v 0 = G in. Since is C 1,1 and L 0 is a second-order elliptic operator with constant coefficients, it is known that v 0 H 2 and v 0 H 2 C G L 2 see [4, Chapter 8] or [10, Chapter 4]. This, together with , gives w ε G dx Cε u 0 H 2 G L 2. By duality we obtain w ε L 2 Cε u 0 H 2. Thus u ε u 0 L 2 Cε u 0 H 2 + εχx/εη εs 2 ε u 0 L 2 which completes the proof. Cε u 0 H 2,

25 Analogous results hold for Neumann problems. Zhongwei Shen 25 Theorem Suppose that A and satisfy the same conditions as in Theorem Let u ε H 1 ε 0 be the weak solution of the Neumann problem Then, if u 0 W 2,q, u ε u 0 L p Cε u 0 W 2,q, where q = p = d+1 2d and C depends only on µ and. Theorem Assume that A is 1-periodic and satisfies Let be a bounded C 1,1 domain in R d. Let u ε H 1 ε 0 be the weak solution of Assume that u 0 H 2. Then u ε u 0 L 2 Cε u 0 H 2, where C depends only on µ and Problems for Section 2 Problem Prove the estimate Problem Show that if g is 1-periodic, gx/ε S ε f L p O Cε 1 g L p Y f L p Oε/2. Problem Prove Proposition Problem Prove Proposition Problem Let u ε be the solution of the Neumann problem with u ε dx = 0, and u 0 the homogenized solution. Let w ε be defined as in Show that the inequality holds for any ψ H 1. Problem Let w ε be the same as in Problem Show that w ε H 1 C ε 2 u 0 L 2 \ ε + ε u 0 L 2 + u 0 L 2 5ε for 0 < ε < 1. Consequently, if u 0 H 2, w ε H 1 C ε u 0 H 2. The constant C depends only on µ and. Problem Suppose that A is 1-periodic and satisfies Also assume that A = A. Let be a bounded Lipschitz domain in R d. Let w ε be the same as in Problem Show that w ε H 1 C ε F L q + g L 2, where q = d+1 2d and C depends only on µ and. 3. Uniform Lipschitz Estimates - Part I In this section we prove uniform Lipschitz estimates, using a compactness method of Avellaneda and Lin [2].

26 26 Quantitative Homogenizationof Elliptic Operators with Periodic Coefficients A very important feature of the family of operators L ε, ε > 0 is the following rescaling property: if L ε u ε = F and vx = uε rx, then L ε r v = G, where Gx = r 2 Frx. It plays an essential role in the compactness method as well as in numerous other rescaling arguments. The property of translation is also important to us: if div Ax/ε u ε = F and v ε x = u ε x x 0, then div Ãx/ε v ε = F, where Ãy = Ay ε 1 x 0 and Fx = Fx x 0. The matrix à is 1-periodic and satisfies the same ellipticity condition as A. It also satisfies the same smoothness condition that we will impose on A, uniformly in ε > 0 and x 0 R d Interior estimates Theorem Assume that the matrix A satisfies and is 1-periodic. Let u ε H 1 B be a weak solution of L ε u ε = F in B, where B = Bx 0, R for some x 0 R d and R > 0, and F L p B for some p > d. Suppose that 0 < ε r R. Then u ε 2 Bx 0,r C p u ε 2 + R Bx 0,R where C p depends only on µ and p. Bx 0,R 1/p F p, Estimate should be regarded as a Lipschitz estimate down to the microscopic scale ε. In fact, under the Hölder continuity condition on A, Ax Ay λ x y τ for any x, y R d, where λ > 0 and τ 0, 1], the full-scale Lipschitz estimate follows from Theorem by a blow-up argument. Theorem Suppose that A satisfies and is 1-periodic. Also assume that A satisfies the smoothness condition Let u ε H 1 B be a weak solution to L ε u ε = F in B for some ball B = Bx 0, R, where F L p B for some p > d. Then u ε x 0 C p u ε 2 + R B where C p depends only on p, µ and λ, τ. 1/p F p, B

27 Zhongwei Shen 27 Proof. We give the proof of Theorem 3.1.2, assuming Theorem By translation and dilation we may assume that x 0 = 0 and R = 1. The ellipticity condition, together with the Hölder continuity assumption 3.1.2, allows us to use the following local regularity result: if divax u = F in B0, 1, where F L p B0, 1 for some p > d, then u0 C p u 2 + B0,1 B0,1 1/p F p, where C p depends only on d, µ, p, λ and τ see e.g. [7]. We may assume that 0 < ε < 1/2, as the case ε 1/2 follows directly from and the observation that the coefficient matrix Ax/ε is uniformly Hölder continuous for ε 1/2. To handle the case 0 < ε < 1/2, we use a blow-up argument and the estimate Let wx = ε 1 u ε εx. Since L 1 w = εfεx in B0, 1, it follows again from that 1/2 w0 C w 2 + B0,1 1/2 C u ε 2 + ε 1 d p B0,ε εfεx p dx B0,1 B0,1 1/p 1/p F p, where C depends only on d, p, µ, λ and τ. This, together with with r = ε and the fact that w0 = u ε 0, gives the estimate Remark Let Then L ε u ε = 0 in R d. Since u ε = x j + εχ j x/ε. u ε = x j + χ j x/ε, no uniform estimate for u ε can be expected beyond Lipschitz unless χ = 0. In the rest of this subsection we will assume that A satisfies and is 1- periodic. No smoothness condition on A is needed. The proof uses only the interior C 1,α estimate for elliptic systems with constant coefficients. Lemma One-step improvement. Let 0 < σ < ρ < 1 and ρ = 1 d p. There exist constants ε 0 0, 1/2 and θ 0, 1/4, depending only on µ, σ and ρ, such

28 28 Quantitative Homogenizationof Elliptic Operators with Periodic Coefficients that u ε x u ε x j + εχ j x/ε u ε 2 dx B0,θ B0,θ x j 1/2 1/p θ 1+σ max u ε 2, F p B0,1 B0,1, B0,θ whenever 0 < ε < ε 0, and u ε H 1 B0, 1 is a weak solution of L ε u ε = F in B0, 1. Proof. Estimate is proved by contradiction, using Theorem and the following observation: for any θ 0, 1/4, u sup ux u x j B0,θ B0,θ x j x θ C θ 1+ρ u C 0,ρ B0,θ C θ 1+ρ u C 0,ρ B0,1/4 1/2 C 0 θ 1+ρ u 2 + B0,1/2 B0,1/2 1/p F p, where u is a solution of diva 0 u = F in B0, 1/2 and A 0 is a constant matrix satisfying the ellipticity condition. We mention that the last inequality in is a standard C 1,ρ estimate for second-order elliptic systems with constant coefficients, and that the constant C 0 depends only on d, µ and ρ [7]. Since σ < ρ, we may choose θ 0, 1/4 so small that 2 d+1 C 0 θ 1+ρ < θ 1+σ. We claim that the estimate holds for this θ and some ε 0 0, 1/2, which depends only on d, µ, σ and ρ. Suppose this is not the case. Then there exist sequences ε l 0, 1/2, A l satisfying and 1.1.3, F l L p B0, 1, and u l H 1 B0, 1, such that ε l 0, div A l x/ε l u l = Fl in B0, 1, and B0,θ u l 2 1, B0,1 u l x 1/p F l p 1, B0,1 B0,θ u l x j + ε l χ l,j x/ε l B0,θ > θ 1+σ, u l x j 2 dx

29 Zhongwei Shen 29 where χ l,j denote the correctors associated with the 1-periodic matrix A l. By and Caccioppoli s inequality, the sequence u l is bounded in H 1 B0, 1/2. By passing to subsequences, we may assume that u l u weakly in L 2 B0, 1, u l u weakly in H 1 B0, 1/2, F l F weakly in L p B0, 1, Â l A 0, where Âl denotes the homogenized matrix for A l. Since p > d, F l F weakly in L p B0, 1 implies that F l F strongly in H 1 B0, 1. It then follows by Theorem that u H 1 B0, 1/2 is a solution of div A 0 u = F in B0, 1/2. We now let l in and This leads to and B0,θ u 2 1, B0,1 ux B0,θ u x j B0,θ 1/p F p 1, B0,1 u 1/2 2 dx θ 1+σ, x j where we have used the fact that u l u strongly in L 2 B0, 1/2. Here we also have used the fact that the sequence χ l,j is bounded in L 2 Y. Finally, we note that by , and , θ 1+σ C 0 θ 1+ρ u 2 + B0,1/2 B0,1/2 1/p F p < 2 d+1 C 0 θ 1+ρ, which is in contradiction with the choice of θ. This completes the proof. Remark Since inf f α 2 = f α R E E for any f L 2 E, we may replace ffl B0,θ u ε in by the average [ u ε x j + ε χ j x/ε ] u ε dx. B0,θ x j B0,θ The observation will be used in the proof of the next lemma. Lemma Iteration. Let 0 < σ < ρ < 1 and ρ = 1 d p. Let ε 0, θ be the constants given by Lemma Suppose that 0 < ε < θ k 1 ε 0 for some k 1 and u ε is a solution of L ε u ε = F in B0, 1. Then there exist constants Eε, l = E j ε, l R d for 1 l k, such that if E f v ε = u ε x j + εχ j x/ε E j ε, l, 2

30 30 Quantitative Homogenizationof Elliptic Operators with Periodic Coefficients then B0,θ l v ε θ l1+σ max 1/2 2 v ε B0,θ l u ε 2, B0,1 B0,1 1/p F p. Moreover, the constants Eε, l satisfy /2 Eε, l C max u ε 2, B0,1 1/2 Eε, l + 1 Eε, l C θ lσ max u ε 2, B0,1 where C depends only on d, µ, σ and ρ. B0,1 1/p F p, B0,1 1/p F p, Proof. We prove by an induction argument on l. The case l = 1 follows from Lemma and Remark 3.1.5, with u ε E j ε, 1 = x j B0,θ set Eε, 0 = 0. Suppose now that the desired constants Eε, l exist for all integers up to some l, where 1 l k 1. To construct Eε, l + 1, consider the function wx = u ε θ l x θ l x j + εχ j θ l x/ε E j ε, l B0,θ l [ uε y j + εχ j y/ε E j ε, l ] dy. By the rescaling property and the equation for correctors, L ε w = F θ l l in B0, 1, where F l x = θ 2l Fθ l x. Since εθ l εθ k+1 ε 0, it follows from Lemma and Remark that B0,θ w x j + εθ l χ j θ l x/ε B0,θ θ 1+σ max [ w B0,θ y j + θ l εχ j θ l y/ε w 2, B0,1 B0,1 w x j B0,θ 1/p F l p. ] w dy 2 dx x j

31 Zhongwei Shen 31 By the induction assumption, w 2 B0, θ l1+σ max u ε 2, B0,1 B0,1 1/p F p. Also, since 0 < ρ = 1 d p, 1/p F l p θ l1+ρ B0,1 Hence, the RHS of is bounded by 1/2 θ l+11+σ max u ε 2, B0,1 with 1/p F p. B0,1 B0,1 Finally, note that the LHS of may be written as u ε x j + εχ j x/ε E j ε, l + 1 B0,θ l+1 B0,θ l+1 By Caccioppoli s inequality, [ uε y j + εχ j y/ε E j ε, l + 1 ] dy E j ε, l + 1 = E j ε, l + θ l Eε, l + 1 Eε, l θ l Cθ l max C θ lσ max w 2, B0,1 u ε 2, B0,1 B0,θ w x j. w 2 B0,θ B0,1 B0,1 1/p F p. 2 1/2 F l p 1/p F p, dx where we have used estimates and for the last inequality. Thus we have established the second inequality in , from which the first follows by summation. This completes the induction argument and thus the proof. Proof of Theorem By translation and dilation we may assume that x 0 = 0 and R = 1. We may also assume that 0 < ε < ε 0 θ, where ε 0, θ are constants given by Lemma The case ε 0 θ ε < 1 is trivial.

32 32 Quantitative Homogenizationof Elliptic Operators with Periodic Coefficients Now suppose that 0 < ε < ε 0 θ. Choose k 2 so that ε 0 θ k ε < ε 0 θ k 1. It follows by Lemma that u ε B0,θ k 1 C θ k 1/2 2 u ε B0,θ k 1 u ε 2 + B0,1 B0,1 1/p F p. This, together with Caccioppoli s inequality, gives u ε 2 B0,ε 0 θ k 1 u ε 2 C B0,ε C θ k C B0,θ k 1 u ε u ε 2 + B0,1 1/2 2 u ε + θ k B0,θ k 1 1/p F p. B0,1 B0,θ k 1 1/2 F 2 By replacing u ε with u ε ffl B0,1 u ε, we obtain u ε 2 B0,ε C B0,1 u ε 1/2 2 u ε + B0,1 B0,1 1/p F p, from which the estimate follows by the Poincaré inequality Boundary Lipschitz estimates for the Dirichlet problem Theorem Suppose that A is 1-periodic and satisfies and Let be a bounded C 1,α domain in R d. Suppose that Lε u ε = F in Bx 0, R, u ε = f on Bx 0, R, where x 0 and 0 < R < diam. Then u ε L Bx 0,R/2 1/2 C u ε 2 + R Bx 0,R F p Bx 0,R + f L Bx 0,R + R σ tan f C 0,σ Bx 0,R, where p > d, σ > 0, and C depends only on d, µ, p, σ, λ, τ and.

33 Zhongwei Shen 33 Theorem 3.2.1, together with the interior Lipschitz estimate, gives the following. Theorem Assume that A and satisfy the same conditions as in Theorem Let u ε be the solution of the Dirichlet problem, L ε u ε = F in and u ε = f on. Then u ε L C F L p + f C 1,σ, where p > d, σ > 0, and C depends only on d, µ, p, σ, λ, τ and. Outlines of the proof of Theorem Step 1. Use the compactness method to establish the boundary Hölder estimate in a C 1 domain: u ε C 0,σ Bx 0,R/2 C R σ u ε 2, Bx 0,R where σ 0, 1, x 0, 0 < R < diam, L ε u ε = 0 in Bx 0, R and u ε = 0 on Bx 0, R. Step 2. Let G ε x, y denote the Green function for L ε in a bounded C 1 domain. Use the Hölder estimate in Step 1 to prove that for any 0 < σ 1, σ 2 < 1, G ε x, y C[δx]σ 1[δy] σ 2 x y d 2+σ 1+σ 2 for any x, y, where δx = distx,. Step 3. Use the estimate for the Green function in Step 2 to show that if L ε u ε = 0 in and u ε = g on, then for 0 < ε < R < diam, u ε 2 C g + ε 1 g L. Bx 0,R Step 4. Let Φ ε x denote the solution of the Dirichlet problem L ε Φε = 0 in and Φε = x on. The vector-valued function Φ ε x is called the Dirichlet corrector for L ε in. Note that if v ε = Φ ε x εχx/ε, then L ε vε = 0 in and vε = εχx/ε on. Use the estimate in Step 3 to show that if is C 1,α, Φ ε L C, where C depends only on d, µ, λ, τ and.

34 34 Quantitative Homogenizationof Elliptic Operators with Periodic Coefficients Step 5. Use the compactness method, together with the Lipschitz estimate for the Dirichlet corrector in Step 4, to prove Theorem A detailed proof may be found in [2, 11] Problems for Section 3 Problem Let be a bounded Lipschitz domain. Let u ε be a weak solution of L ε u ε = F + divf in Bx 0, r with u ε = g on Bx 0, r, where x 0 and 0 < r < r 0. Show that u ε 2 dx C r 2 u ε 2 dx + Cr 2 F 2 dx Bx 0,r/2 Bx 0,r + C f 2 dx + C Bx 0,r + C r 2 G 2 dx, Bx 0,r where G H 1 Bx 0, r and G = g on Bx 0, r. Bx 0,r Bx 0,r 0 G 2 dx Problem Prove Theorem 3.2.2, using Theorem and interior estimates. Problem Use the compactness method to prove the boundary Hölder estimate. 4. Uniform Lipschitz estimates - Part II In this section we prove uniform Lipschitz estimates for the Neumann problem, L ε u ε = F in, u ε = g on, ν ε where L ε defined by = div Ax/ε and u ε ν ε denotes the conormal derivative of u ε, u ε ν ε = n i a ij x/ε u ε x j. Our approach is based on a general scheme for establishing large-scale regularity estimates in homogenization, developed by S. N. Armstrong and C. Smart [1] in the study of stochastic homogenization. Roughly speaking, the scheme states that if a function u ε is well approximated by C 1,α functions at every scale greater than ε, then u ε is Lipschitz continuous at every scale greater than ε. The approach relies on a very weak result on convergence rates and does not involve correctors in a direct manner. In comparison with the compactness method used in Section 3, when applied to boundary Lipschitz estimates, it does not require a-priori Lipschitz estimate for boundary correctors.

35 4.1. Approximation of solutions at the large scale Let Zhongwei Shen 35 D r = Dr, ψ = x, ψx : x < r and ψx < x d < ψx + 10M 0 + 1r, r = r, ψ = x, ψx : x < r, where ψ is a Lipschitz function in R d 1 such that ψ0 = 0 and ψ M 0. Theorem Suppose that A is 1-periodic and satisfies Let u ε H 1 D 2r be a weak solution to L ε u ε = F in D 2r and u ε = g ν ε on 2r, where F L 2 D 2r and g L 2 2r. Assume that r ε. Then there exists w H 1 D r such that L 0 w = F in D r and w = g ν 0 on r, and u ε w 2 D r r 2 ε α C r D 2r u ε 2 F 2 + r D 2r where C > 0 and α 0, 1/2 depend only on µ and M 0. 1/2 g 2, 2r Lemma Let = D r for some 1 r 2. Let u ε H 1 ε 0 be a weak solution to the Neumann problem, L ε u ε = F in and u ε ν ε = g on, where F L 2, g L 2, and F dx + g dσ = 0. Then there exists p > 2, depending only on µ and M 0, such that u ε L p C F L 2 + g L 2, where C > 0 depend only on µ and M 0. Proof. Consider the Neumann problem: L ε v ε = divf in and v ε ν ε = 0 on, where f C 1 0 ; Rd. Clearly, v ε L 2 C f L 2. Using the reverse Hölder inequality and the real-variable argument in Section 5, one may deduce that v ε L p C f L p, for any 2 < p < p. Since this is also true for the adjoint operator L ε, by a duality argument, it follows that u ε L p C F L q + g B 1/p,p,

36 36 Quantitative Homogenizationof Elliptic Operators with Periodic Coefficients for 2 < p < p, where q 1 = p 1 + d 1 and B 1/p,p denotes the dual of the Besov space B 1/p,p. By choosing p 2, p close to 2 so that L 2 L q and L 2 B 1/p,p, we see that the estimate follows from Lemma Let u ε H 1 ε 0 be the weak solution to the Neumann problem with u ε dx = 0, where = D r for some 1 r 2. Then u ε u 0 L 2 Cεσ F L 2 + g L 2 for any 0 < ε < 2, where σ > 0 and C > 0 depend only on µ and M 0. Proof. It follows from that u ε u 0 L 2 ε C 2 u 0 L 2 \ ε + ε u 0 L 2 + u 0 L 2 5ε, where t = x : distx, < t. To bound the RHS of 4.1.7, we first use interior estimates for L 0 to obtain 2 u 0 2 C Bx,δx/8 [δx] 2 u C Bx,δx/4 F 2 Bx,δx/4 for any x, where δx = distx,. We then integrate both sides of the inequality above over the set \ ε. Observe that if x y < δx/4, then δx δy x y < δx/4, which gives It follows that 4/5δy < δx < 4/3δy. 2 u 0 y 2 dy \ ε C [δy] 2 u 0 y 2 dy + C \ ε/2 Fy 2 dy C u 0 2s dy 1/s [δy] 2s dy \ ε/2 Cε 1 s 1 u 0 2 L 2s + +C Fy 2 dy, 1/s + C Fy 2 dy where s > 1 and we have used Hölder s inequality for the second step. p = 2s > 2. We see that Let ε 2 u 0 L 2 \ ε Cε 1 2 p 1 u 0 L p + Cε F L 2 Cε 1 2 p 1 F L 2 + g L 2, where we have used for the last inequality. Also, by Hölder s inequality, u 0 L 2 5ε Cε 2 1 p 1 u 0 L p Cε 1 2 p 1 F L 2 + g L 2. The inequality with σ = p > 0