# A New Estimate for the Homogenization Method for Second-Order Elliptic Problem with Rapidly Oscillating Periodic Coefficients

Authors: Xiong Liu; Wenming He
Journal: Journal of Function Spaces
 (2021)

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## Abstract

In this paper, we will investigate a multiscale homogenization theory for a second-order elliptic problem with rapidly oscillating periodic coefficients of the form∂/∂xiaijx/ε,x∂uεx/∂xj=fx. Noticing the fact that the classic homogenization theory presented by Oleinik has a high demand for the smoothness of the homogenization solution u0, we present a new estimate for the homogenization method under the weaker smoothness that homogenization solution u0 satisfies than the classical homogenization theory needs.


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## Body

## 1. Introduction

Many people investigated the second-order elliptic problem with a fixed boundary. As far as we know, there is not any work related to the elliptic problem with periodic boundary (see [1–3]). In this article, we will consider the following multiscale elliptic model problem:
(1)Lεuεx≡∂∂xiaijxε,x∂uε∂xj=fx,inΩ,uεx=gx,on∂Ω.Here,Ω⊂Rnn≥1 is a bounded domain, and the matrix of coefficients aijξ,x:Rn⟶Rn×n is symmetric and satisfies the following conditions:
(2)γξ2≤aijξ,xξiξj≤γ−1ξ2,ξ∈Rn,forsomeγ∈0,1aijξ+ξ′=aijξ,ξ∈Rn,ξ′∈Zn,1≤i,j≤nAssume thatQ=0,1n. By the homogenization method, Oleinik et al. (see [4, 5]) obtained the 1-order approximation u~x of uε as follows:
(3)u~x=u0x+εNkx,xε∂u0x∂xk,where Nkx,ξ is a 1-periodic function and satisfies the following equations:
(4)∂∂ξiaijx,ξ∂Nkx,ξ∂ξj=−∂aikx,ξ∂ξi,inRn,∫QNkξ,xdξ=0,a∧ijx=∫Qaijx,ξ+aikx,ξ∂Njx,ξ∂ξkdξ,and the homogenization solution u0 satisfies the problem as follows:
(5)L0u0x≡∂∂xia∧ijx∂u0∂xj=fx,inΩ,u0x=gx,on∂Ω.Oleinik et al. (see [5], p. 28) proved the following result.We end this section with the details of some notations. Throughout this paper, the Einstein summation convention is used: summation is taken over repeated indices, andρx,∂Ω denotes the distance between x and ∂Ω.


## 2. Some Useful Lemmas

Lemma 1.
Under the assumption thatu0∈H2Ω, there holds
(6)uε−u~H1Ω≤cε1/2u0H2Ω.There are numerous literatures discussing the homogenization method (see [1, 2, 4–9]). There also are many works (see [3, 10–16]) discussing the numerical methods of the multiscale homogenization problem. We observe that most of them are based on the assumption u0∈H2Ω, which is unrealistic for some problems. For example, when f∉L2Ω. Let u~ix=∂u0x/∂xi+∂Nkξ,x/∂ξi∂u0x/∂xk. As far as we know, it is the first time for us to estimate u~ix−∂uεx/∂xi under the assumption that the homogenization solution u0 belongs to the Sobolev space H1+sΩ for the case that 0<s<1.Lemma 2.
Assume thatu∈H1+sΩ∩W1,∞K2r. Then,
(7)∇u−urL2Ω≤crsuH1+sΩ+r1/2∇uL∞K2r,(8)∇2urL2Ω≤crs−1uH1+sΩ+r−1/2∇uL∞K2r.Proof.
One observes that∇u−urL2Ω2 can be split into
(9)∇u−urL2Ω2=∇u−urL2Ω\Kr2+∇u−urL2Kr2.
We first estimate∇u−urL2Ω\Kr2. Assume that x∈Ω\Kr and Bx,r=y∈Ω:x−y≤r. Note that the definition of ωrz implies ∫Ωωrx−ydy=1. By the definitions of ωrz and urx, we have, for any 1≤i≤n,
(10)∂urx∂xi=∫Ω∂ωrx−y∂xiuydy=−∫Ω∂ωrx−y∂yiuydy=∫Ωωrx−y∂uy∂yidy=∫Bx,rωrx−y∂uy∂yidy.
Using (10), we obtain
(11)∂urx∂xi−∂ux∂xi=∫Bx,rωrx−y∂uy∂yi−∂ux∂xidy.
Furthermore, from the definition ofωrz and (11), it follows that
(12)∂u−ur∂xiL2Ω\Kr2=∫Ω\Kr∫Bx,rωrx−y∂uy∂yi−∂ux∂xidy2dx,(13)≤cr−2n∫Ω\Kr∫Bx,r∂uy∂yi−∂ux∂xidy2dx.
Note that(14)∫Bx,r∂uy∂yi−∂ux∂xidy≤crs+n/2∫Bx,r∂uy∂yi−∂ux∂xix−y−s−n/2dy.
This, together with (13), gives
(15)∂u−ur∂xiL2Ω\Kr2≤cr2s−n∫Ω\Kr∫Bx,r∂uy∂yi−∂ux∂xix−y−s−n/2dy2dx≤cr2s∫Ω\Kr∫Bx,r∂uy∂yi−∂ux∂xi2x−y−2s−ndydx≤cr2suH1+sΩ2.
Next we estimate∇u−urL2Kr2. Assume that x∈Kr. Set ω¯rx−y=ωrx−y/∫Bx,rωrx−ydy. Let x¯=x or x+Δx. We have
(16)urx¯=∫Bx¯,rω¯rx¯−yuydy=ux¯+∫Bx¯,rω¯rx¯−yuy−uxdy.
Let∣Δx∣≤r. Note that ω¯rz=0 whenever ∣z∣≥r. By (16), one observes that urx+Δx−urx can be decomposed into
(17)urx+Δx−urx=∫Bx+Δx,rω¯rx+Δx−yuy−uxdy−∫Bx,rω¯rx−yuy−uxdy=∫Bx+Δx,r−∫Bx,rω¯rx+Δx−yuy−uxdy+∫Bx,rω¯rx+Δx−y−ω¯rx−yuy−uxdy=I1+I2.
We need estimatesI1 and I2. Assume that y∈Bx+Δx,r\Bx,r. Note that x+Δx∈Ω. One observes that ∫Bx+Δx,r/2‍dy≥crn. Then, we have
(18)∫Bx+Δx,r/2ωrx−ydy≥c.
By the definition ofω¯rx and (18), we have
(19)ω¯rx+Δx−y≤cr−n.
Note that(20)uy−ux≤cruW1,∞K2r+Δx.
Inserting (19) and (20) into (17), we have
(21)I1≤crn−1Δxr−nruW1,∞K2r+Δx≤cΔxuW1,∞K2r+Δx.
We turn now to the estimation ofI2. We split ω¯rx+Δx−y−ω¯rx−y into
(22)ω¯rx+Δx−y−ω¯rx−y=ωrx+Δx−y∫Bx+Δx,rωrx+Δx−ydy−ωrx−y∫Bx,rωrx−ydy=ωrx+Δx−y∫Bx+Δx,rωrx+Δx−ydy−1−∫Bx,rωrx−ydy−1+ωrx+Δx−y−ωrx−y∫Bx,rωrx−ydy=J1+J2.
We need to estimate the two items of the right-hand side of (22). Note that
(23)∫Bx+Δx,rωrx+Δx−ydy−∫Bx,rωrx−ydy≤∫Bx+Δx,r−∫Bx,rωrx+Δx−ydy+∫Bx,rωrx+Δx−y−ωrx−ydy∣≤crn−1Δxr−n+crnr−n−1Δx≤cr−1Δx.
By (22) and (23), we have
(24)J1≤cr−ncr−1Δx≤cr−n−1Δx.
To estimateJ2, we have
(25)J2≤cr−1Δxr−n≤cr−n−1Δx.
Plugging the above two estimates into (22), we obtain
(26)ω¯rx+Δx−y−ω¯rx−y≤cr−n−1Δx.
This, together with (17), gives
(27)I2≤crnr−n−1ΔxruW1,∞K2r≤cΔxuW1,∞K2r.
Inserting (21) and (27) into (17), we have
(28)urx+Δx−urx≤cΔxuW1,∞K2r+Δx.
Furthermore, letΔx⟶0, we have
(29)urW1,∞Kr≤cuW1,∞K2r,where we have used (28). Then, (7) follows by combining (15) and (29). We turn now to the estimation of ∇2urL2Ω. We decompose ∇2urL2Ω into
(30)∇2urL2Ω2=∇2urL2Ω\Kr2+∇2urL2Kr2.
We first estimate∇2urL2Ω\Kr2. Assume that x∈Ω\Kr. By (10), we have, for any 1≤i,j≤n,
(31)∂2urx∂xi∂xj=∂∫Ωωrx−y∂uy/∂yidy∂xj=∫Ω∂ωrx−y∂xj∂uy∂yidy.
Note thatx∈Ω\Kr. By the definition of ωrx−y, we have ∫Ω∂ωrx−y/∂xjdy=0. Then, by (28) and (31), we have
(32)∂2urx∂xi∂xj=∂∫Ωωrx−y∂uy/∂yidy∂xj=∫Ω∂ωrx−y∂xj∂uy∂yi−∂ux∂xidy.
Finally, similarly to (15), by (32), we have
(33)∇2urL2Ω\Kr≤crs−1uH1+sΩ.
We turn now to the estimation of∇2urL2Kr. Similarly to (17), we have
(34)urx+2Δx−2urx+Δx+urx=∫Bx,2rω¯rx+2Δx−y−2ω¯rx+Δx−y+ω¯rx−yuy−uxdy.
Note that the definition ofωrz implies ω¯rW2,1Rn≤cr−2. Therefore, let Δx⟶0, from (34), it follows that
(35)urW2,∞Kr≤cω¯rW2,1RncruW1,∞K2r≤cr−2ruW1,∞K2r≤cr−1uW1,∞K2r.
The desired result (8) follows by combining (33) and (35).☐


## 3. A New Estimate for Multiscale Homogenization Method

In this section, we give the main results as follows.Theorem 3.
Assume thatKr=x∈Ω∣ρx,∂Ω≤r and Q=0,1n. Assume also that Nk∈W1,∞Q and u0∈H1+sΩ∩W1,∞Kε for some 0<s<1. Then,
(36)∂uε∂xi−u~iL2Ω≤cε1/2u0W1,∞Kε+εsu0H1+sΩ.Assume thatχz∈C∞Rn is the cutoff function satisfying 0≤χz≤1, and χz=1 if z≤1/2, and χz=0 if z≥1. Let ωrz=χz/r/∫Rnχy/rdy. One observes that ∫B0,rωrzdz=1 and ωrWk,∞Rn≤cr−k−n for all k≥0. Set urx=∫Ωωrx−yuydy/∫Ωωrx−ydy. In the process of proving Theorem 3, we need the above Lemma 2.Based on Lemma2, we can prove Theorem 3 as follows:Proof.
Assume thatωrz is defined as in Lemma 2. Set
(37)u¯r0x=∫Ωωrx−yu0ydy∫Ωωrx−ydy,frx=∂∂xja^ijx∂u¯r0x∂xi.
We introduceurεx by the following problem:
(38)Lεurεx=frx,inΩ,urεx=u¯r0x,on∂Ω.
One observes thatur0x and u~rx are the homogenization solution of (38) and the 1-order approximation of urεx, respectively. We decompose ∂uεx/∂xi−u~ix into
(39)∂uεx∂xi−u~ix=∂uε−urεx∂xi+∂urε−u~rx∂xi+∂u~rx∂xi−u~ix.
We first estimate∇uε−urεx. Let B1εx=uε−urεx. Note that B1εx satisfies the following problem:
(40)LεB1εx=fx−frx,x∈Ω,B1εx=gx−u¯r0x,x∈∂Ω.
One observes thatB1εx can be split into
(41)B1εx=e1x+e2εx,where e1x=u0−u¯r0x and e2εx satisfies the following problem:
(42)Lεe2εx=f−frx−∂∂xiaijx,xε∂e1x∂xj,inΩ,e2εx=0,on∂Ω.
From the combination of the definition ofu¯r0 and (7), it follows that
(43)∇u¯r0−u0L2Ω≤crsu0H1+sΩ+r1/2u0W1,∞K2r.
To estimatee2εx, by (8) and the definitions of u¯r0x and frx, one observes that
(44)∇e2εL2Ω≤f−frH−1Ω+∇u¯r0−u0L2Ω≤c∇u¯r0−u0L2Ω≤crsu0H1+sΩ+cr1/2u0W1,∞K2r.
Combining (39), (41), (43), and (44), we have
(45)∇uε−urεL2Ω=∇B1εL2Ω≤crsu0H1+sΩ+r1/2u0W1,∞K2r.
Next, we estimateurε−u~rx. Set B2εx=urε−u~rx. By the method of asymptotic expansion (see [7], p. 27), one finds that B2εx can be split into B2εx=wrεx+θrεx, where wrεx and θrεx are defined by
(46)Lεwrεx≡∂∂xiaijx,xε∂wrεx∂xj=∂Fr,ix∂xi,inΩ,wrεx=0,on∂Ω,Lεθrεx=0,inΩ,θrεx=−εNkxε,x∂u¯r0x∂xk,on∂Ω,respectively, where
(47)Fr,ix=−aijx,xε+aikx,xε∂Njξ∂ξk−a^ij∂u¯r0x∂xj+εaijx,xεNkxε∂2u¯r0x∂xj∂xk.
We first estimatewrεx. Note that (8) implies
(48)u¯r0H2Ω≤cr−1+su0H1+sΩ+r−1/2u0W1,∞K2r.
By the method of asymptotic expansion (see ([7], p. 27), from (48), it follows that
(49)wrεH1Ω≤cεu¯r0H2Ω≤cεr−1+su0H1+sΩ+εr−1/2u0W1,∞K2r.
Assume thatϕx∈C∞Ω is a cutoff function satisfying ϕx=1 if ρx,∂Ω≤ε, and ϕx=0 if ρx,∂Ω≥2ε, and ∇ϕL∞Ω≤c2ε−1. We split θrεx into
(50)θrεx=ψrεx+ψ^rεx,where ψrεx=−εNkx/ε,x∂u¯r0x/∂xkϕx and ψ^rεx satisfies the following problem:
(51)Lεψ^rεx=−∂∂xiaijx,xε∂ψrx∂xj,x∈Ω,ψ^rεx=0,x∈∂Ω.
To estimateψrεx, one has
(52)ψrεH1Ω≤cεr−1u¯r0H1Kε+cεu¯r0H2Kε.
We now estimateu¯r0H1Kε. Assume that r=ε and VKε denotes the volume of Kε if n=3, or the area of Kε if n=2. One observes that
(53)u¯r0H1Kε≤cVKεu¯r0W1,∞Kε≤cε1/2u¯r0W1,∞Kε≤cε1/2u0W1,∞Kε.
The combination of (8), (52), and (53) gives
(54)ψrεH1Ω≤cε1/2u0W1,∞K2ε+εsu0H1+sΩ.
Using (51) and (54), we derive
(55)ψ^rεH1Ω≤cψrεH1Ω≤cε1/2u0W1,∞K2ε+εsu0H1+sΩ.
The above two estimates, together with (50), imply
(56)θrεH1Ω≤cε1/2+ε3/2r−1u0W1,∞K2ε.
Furthermore, by (49) and (56), we have
(57)urε−u~rH1Ω=B2εH1Ω≤cε1/2+ε3/2ε−1u0W1,∞K2ε+εε−1+su0H1+sΩ≤cε1/2u0W1,∞K2ε+εsu0H1+sΩ,
We next estimate∂u~rx/∂xi−u~ix. Assume that r=ε. Note that the definitions of ur0 and urε imply ur0x=u¯r0x. By (7) and (47), we have
(58)∂u~r∂xi−u~iL2Ω≤c∇u0−u¯r0L2Ω≤cεsu0H1+sΩ+ε1/2u0W1,∞K2ε.
Assume thatr=ε. This, together with (39), (45), and (57), gives the desired result (36).☐



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