JFSJournal of Function Spaces2314-88882314-8896Hindawi10.1155/2021/80368148036814Research ArticleA New Estimate for the Homogenization Method for Second-Order Elliptic Problem with Rapidly Oscillating Periodic Coefficientshttps://orcid.org/0000-0002-5452-653XLiuXiongHeWenmingChenChuanjunSchool of Mathematics and StatisticsLingnan Normal UniversityZhanjiangGuangdong 524048Chinalingnan.edu.cn2021196202120212942021146202119620212021Copyright © 2021 Xiong Liu and He Wenming.This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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,xuεx/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.

Lingnan Normal UniversityZL1810National Natural Science Foundation of China11671304
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 ). In this article, we will consider the following multiscale elliptic model problem: (1)Lεuεxxiaijxε,xuεxj=fx,inΩ,uεx=gx,onΩ.

Here, ΩRnn1 is a bounded domain, and the matrix of coefficients aijξ,x: RnRn×n is symmetric and satisfies the following conditions: (2)γξ2aijξ,xξiξjγ1ξ2,ξRn,forsomeγ0,1aijξ+ξ=aijξ,ξRn,ξZn,1i,jn

Assume that Q=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,xεu0xxk,where Nkx,ξ is a 1-periodic function and satisfies the following equations: (4)ξiaijx,ξNkx,ξξj=aikx,ξξi,inRn,QNkξ,xdξ=0,aijx=Qaijx,ξ+aikx,ξNjx,ξξkdξ,and the homogenization solution u0 satisfies the problem as follows: (5)L0u0xxiaijxu0xj=fx,inΩ,u0x=gx,onΩ.

Oleinik et al. (see , 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, and ρx,Ω denotes the distance between x and Ω.

2. Some Useful LemmasLemma 1.

Under the assumption that u0H2Ω, there holds (6)uεu~H1Ωcε1/2u0H2Ω.

There are numerous literatures discussing the homogenization method (see [1, 2, 49]). There also are many works (see [3, 1016]) discussing the numerical methods of the multiscale homogenization problem. We observe that most of them are based on the assumption u0H2Ω, which is unrealistic for some problems. For example, when fL2Ω. Let u~ix=u0x/xi+Nkξ,x/ξiu0x/xk. As far as we know, it is the first time for us to estimate u~ixuεx/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 that uH1+sΩW1,K2r. Then, (7)uurL2ΩcrsuH1+sΩ+r1/2uLK2r,(8)2urL2Ωcrs1uH1+sΩ+r1/2uLK2r.

Proof.

One observes that uurL2Ω2 can be split into (9)uurL2Ω2=uurL2Ω\Kr2+uurL2Kr2.

We first estimate uurL2Ω\Kr2. Assume that xΩ\Kr and Bx,r=yΩ:xyr. Note that the definition of ωrz implies Ωωrxydy=1. By the definitions of ωrz and urx, we have, for any 1in, (10)urxxi=Ωωrxyxiuydy=Ωωrxyyiuydy=Ωωrxyuyyidy=Bx,rωrxyuyyidy.

Using (10), we obtain (11)urxxiuxxi=Bx,rωrxyuyyiuxxidy.

Furthermore, from the definition of ωrz and (11), it follows that (12)uurxiL2Ω\Kr2=Ω\KrBx,rωrxyuyyiuxxidy2dx,(13)cr2nΩ\KrBx,ruyyiuxxidy2dx.

Note that (14)Bx,ruyyiuxxidycrs+n/2Bx,ruyyiuxxixysn/2dy.

This, together with (13), gives (15)uurxiL2Ω\Kr2cr2snΩ\KrBx,ruyyiuxxixysn/2dy2dxcr2sΩ\KrBx,ruyyiuxxi2xy2sndydxcr2suH1+sΩ2.

Next we estimate uurL2Kr2. Assume that xKr. Set ω¯rxy=ωrxy/Bx,rωrxydy. Let x¯=x or x+Δx. We have (16)urx¯=Bx¯,rω¯rx¯yuydy=ux¯+Bx¯,rω¯rx¯yuyuxdy.

Let Δxr. Note that ω¯rz=0 whenever zr. By (16), one observes that urx+Δxurx can be decomposed into (17)urx+Δxurx=Bx+Δx,rω¯rx+ΔxyuyuxdyBx,rω¯rxyuyuxdy=Bx+Δx,rBx,rω¯rx+Δxyuyuxdy+Bx,rω¯rx+Δxyω¯rxyuyuxdy=I1+I2.

We need estimates I1 and I2. Assume that yBx+Δx,r\Bx,r. Note that x+ΔxΩ. One observes that Bx+Δx,r/2dycrn. Then, we have (18)Bx+Δx,r/2ωrxydyc.

By the definition of ω¯rx and (18), we have (19)ω¯rx+Δxycrn.

Note that (20)uyuxcruW1,K2r+Δx.

Inserting (19) and (20) into (17), we have (21)I1crn1ΔxrnruW1,K2r+ΔxcΔxuW1,K2r+Δx.

We turn now to the estimation of I2. We split ω¯rx+Δxyω¯rxy into (22)ω¯rx+Δxyω¯rxy=ωrx+ΔxyBx+Δx,rωrx+ΔxydyωrxyBx,rωrxydy=ωrx+ΔxyBx+Δx,rωrx+Δxydy1Bx,rωrxydy1+ωrx+ΔxyωrxyBx,rωrxydy=J1+J2.

We need to estimate the two items of the right-hand side of (22). Note that (23)Bx+Δx,rωrx+ΔxydyBx,rωrxydyBx+Δx,rBx,rωrx+Δxydy+Bx,rωrx+Δxyωrxydycrn1Δxrn+crnrn1Δxcr1Δx.

By (22) and (23), we have (24)J1crncr1Δxcrn1Δx.

To estimate J2, we have (25)J2cr1Δxrncrn1Δx.

Plugging the above two estimates into (22), we obtain (26)ω¯rx+Δxyω¯rxycrn1Δx.

This, together with (17), gives (27)I2crnrn1ΔxruW1,K2rcΔxuW1,K2r.

Inserting (21) and (27) into (17), we have (28)urx+ΔxurxcΔxuW1,K2r+Δx.

Furthermore, let Δx0, we have (29)urW1,KrcuW1,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 xΩ\Kr. By (10), we have, for any 1i,jn, (31)2urxxixj=Ωωrxyuy/yidyxj=Ωωrxyxjuyyidy.

Note that xΩ\Kr. By the definition of ωrxy, we have Ωωrxy/xjdy=0. Then, by (28) and (31), we have (32)2urxxixj=Ωωrxyuy/yidyxj=Ωωrxyxjuyyiuxxidy.

Finally, similarly to (15), by (32), we have (33)2urL2Ω\Krcrs1uH1+sΩ.

We turn now to the estimation of 2urL2Kr. Similarly to (17), we have (34)urx+2Δx2urx+Δx+urx=Bx,2rω¯rx+2Δxy2ω¯rx+Δxy+ω¯rxyuyuxdy.

Note that the definition of ωrz implies ω¯rW2,1Rncr2. Therefore, let Δx0, from (34), it follows that (35)urW2,Krcω¯rW2,1RncruW1,K2rcr2ruW1,K2rcr1uW1,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 that Kr=xΩρx,Ωr and Q=0,1n. Assume also that NkW1,Q and u0H1+sΩW1,Kε for some 0<s<1. Then, (36)uεxiu~iL2Ωcε1/2u0W1,Kε+εsu0H1+sΩ.

Assume that χzCRn is the cutoff function satisfying 0χz1, and χz=1 if z1/2, and χz=0 if z1. Let ωrz=χz/r/Rnχy/rdy. One observes that B0,rωrzdz=1 and ωrWk,Rncrkn for all k0. Set urx=Ωωrxyuydy/Ωωrxydy. In the process of proving Theorem 3, we need the above Lemma 2.

Based on Lemma 2, we can prove Theorem 3 as follows:

Proof.

Assume that ωrz is defined as in Lemma 2. Set (37)u¯r0x=Ωωrxyu0ydyΩωrxydy,frx=xja^ijxu¯r0xxi.

We introduce urεx by the following problem: (38)Lεurεx=frx,inΩ,urεx=u¯r0x,onΩ.

One observes that ur0x and u~rx are the homogenization solution of (38) and the 1-order approximation of urεx, respectively. We decompose uεx/xiu~ix into (39)uεxxiu~ix=uεurεxxi+urεu~rxxi+u~rxxiu~ix.

We first estimate uεurεx. Let B1εx=uεurεx. Note that B1εx satisfies the following problem: (40)LεB1εx=fxfrx,xΩ,B1εx=gxu¯r0x,xΩ.

One observes that B1εx can be split into (41)B1εx=e1x+e2εx,where e1x=u0u¯r0x and e2εx satisfies the following problem: (42)Lεe2εx=ffrxxiaijx,xεe1xxj,inΩ,e2εx=0,onΩ.

From the combination of the definition of u¯r0 and (7), it follows that (43)u¯r0u0L2Ωcrsu0H1+sΩ+r1/2u0W1,K2r.

To estimate e2εx, by (8) and the definitions of u¯r0x and frx, one observes that (44)e2εL2ΩffrH1Ω+u¯r0u0L2Ωcu¯r0u0L2Ωcrsu0H1+sΩ+cr1/2u0W1,K2r.

Combining (39), (41), (43), and (44), we have (45)uεurεL2Ω=B1εL2Ωcrsu0H1+sΩ+r1/2u0W1,K2r.

Next, we estimate urεu~rx. Set B2εx=urεu~rx. By the method of asymptotic expansion (see , p. 27), one finds that B2εx can be split into B2εx=wrεx+θrεx, where wrεx and θrεx are defined by (46)Lεwrεxxiaijx,xεwrεxxj=Fr,ixxi,inΩ,wrεx=0,onΩ,Lεθrεx=0,inΩ,θrεx=εNkxε,xu¯r0xxk,onΩ,respectively, where (47)Fr,ix=aijx,xε+aikx,xεNjξξka^iju¯r0xxj+εaijx,xεNkxε2u¯r0xxjxk.

We first estimate wrεx. Note that (8) implies (48)u¯r0H2Ωcr1+su0H1+sΩ+r1/2u0W1,K2r.

By the method of asymptotic expansion (see (, p. 27), from (48), it follows that (49)wrεH1Ωcεu¯r0H2Ωcεr1+su0H1+sΩ+εr1/2u0W1,K2r.

Assume that ϕxCΩ is a cutoff function satisfying ϕx=1 if ρx,Ωε, and ϕx=0 if ρx,Ω2ε, and ϕLΩc2ε1. We split θrεx into (50)θrεx=ψrεx+ψ^rεx,where ψrεx=εNkx/ε,xu¯r0x/xkϕx and ψ^rεx satisfies the following problem: (51)Lεψ^rεx=xiaijx,xεψrxxj,xΩ,ψ^rεx=0,xΩ.

To estimate ψrεx, one has (52)ψrεH1Ωcεr1u¯r0H1Kε+cεu¯r0H2Kε.

We now estimate u¯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¯r0H1KεcVKεu¯r0W1,Kεcε1/2u¯r0W1,Kεcε1/2u0W1,Kε.

The combination of (8), (52), and (53) gives (54)ψrεH1Ωcε1/2u0W1,K2ε+εsu0H1+sΩ.

Using (51) and (54), we derive (55)ψ^rεH1ΩcψrεH1Ωcε1/2u0W1,K2ε+εsu0H1+sΩ.

The above two estimates, together with (50), imply (56)θrεH1Ωcε1/2+ε3/2r1u0W1,K2ε.

Furthermore, by (49) and (56), we have (57)urεu~rH1Ω=B2εH1Ωcε1/2+ε3/2ε1u0W1,K2ε+εε1+su0H1+sΩcε1/2u0W1,K2ε+εsu0H1+sΩ,

We next estimate u~rx/xiu~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~rxiu~iL2Ωcu0u¯r0L2Ωcεsu0H1+sΩ+ε1/2u0W1,K2ε.

Assume that r=ε. This, together with (39), (45), and (57), gives the desired result (36).☐

Data Availability

The paper’s data available through the email liuxiong980211@163.com or from the author’s ORCID: 0000-0002-5452-653X, and other data is given to the journal of functional spaces https://orcid.org/0000-0002-5452-653X.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

Acknowledgments

This paper is supported by the National Natural Science Foundation of China under grants (No. 11671304) and the PhD Start-up Fund of Lingnan Normal University (No. ZL1810).

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