具有两个m-凸绝缘体和超导体内含物的电导问题的梯度估计.pdf

应用数学MATHEMATICA APPLICATA2024,37(2):540-546具有两个m-凸绝缘体和超导体内含物的电导问题的梯度估计包圆,马飞遥(宁波大学数学与统计学院,浙江 宁波 315211)摘要：本文研究复合材料中的一个内含物的电导率为超导,另一个内含物的电导率为绝缘情况下的电导率问题的梯度估计.本文将内含物的形状从2-凸推广到更一般的情形,得到了当内含物的形状为m-凸(m 2)时解的梯度的上界.关键词：电导率问题;梯度估计;超导;绝缘;m-凸中图分类号：O175.2AMS(2010)主题分类：35J25;74E30文献标识码：A文章编号：1001-9847(2024)02-0540-071.引言由于电导率问题可以作为多种物理现象的数学模型,这一类问题成为了科学和工程领域的一个重要课题,相关的理论已经得到了广泛的研究,并取得了很多研究成果13,712.本文在包含两个内含物的复合材料中研究电导率问题,用边界为C2,(0 0,有?aij?C(e),aij(x)ij|2,Rn,x e.2.预备知识及引理在证明本文的结果之前,我们需要引入一些符号.设x=(x,xn)(x Rn1),通过选取合适的坐标系,我们定义D1和D2之间的狭窄区域如下:r:=(x,xn)e|2+g(x)xn2+f(x),|x|r.令+=r D1=xn=2+f(x),|x|r,=r D2=xn=2+g(x),|x|r,其中f,g C2,().于是本文可以利用函数f和g来刻画内含物的m-凸性,即对于任意的0|x|0.(2.3)定义检验函数 C满足|3=0,其中3=r(+).因此在狭窄区域r上有下式成立:0=ri(aijju)=raijju(x)i+aijju(x)i+aijju(x)i+3aijju(x)i=raijju(x)i+aijju(x)i.从而方程(1.1)的弱形式可以写成raijju(x)i+aijju(x)i=0,C,|3=0,u=C,在+(r)上,aijju(x)i=0,在(r)上.(2.4)542应用数学2024为了处理上界估计中最重要的一步,本文在狭窄区域上引入“翻转延拓”技巧.对任意的 k Z,令Qk:=(z,zn)Rn|z|1,(2k 1)zn(2k+1),并且定义区域Qk的边界如下:k+:=(z,zn)Rn|z|1,zn=(2k+1),(2.5)k:=(z,zn)Rn|z|1,zn=(2k 1),以及k3=Qk(k k+).于是在区域Q0上,方程(2.4)可以写成下列形式:Q0bijzjwzi+0+bnjzjw=0,C(Q0),?03=0,w=C1,在 0+上,bnjzjw=0,在 0上,(2.6)其中bij(z)C(Q0)(0 0 使得?bij(z)?C(Q0)1,bij(z)ij 1|2,Rn,z Q0.(2.7)引理15设w H1(Q0)L(Q0)是方程(2.6)的一个弱解,那么存在只依赖于n,1,1的常数C 0,使得wL(Q0(12)CwL(Q0),其中Q0(12):=z Rn?|z|12,|zn|.证定义Q:=z Rn?|z|1,1 zn 1.考虑Q上的一个奇偶交错延拓,将w延拓为e w,即对k Z有e w(z)=w2k+1(z,zn)=2w2k(z,(4l+1)w2k(z,2(4l+1)zn),k 0;w2k+2(z,zn)=w2k+1(z,2(4l+3)zn),k 0;w2k1(z,zn)=w2k(z,2(4l 1)zn),k 0;w2k2(z,zn)=2w2k1(z,(4l 3)w2k1(z,2(4l 3)zn),k 0.(2.8)通过上述奇偶交错延拓,可知e w可被区域Q上的解w0迭代定义,并满足下列椭圆方程:Qkbijzje wzi+k1bnjzje w=0,C(Qk),?k3=0,e w=C1,在 k+上,bnjzje w=0,在 k上.我们定义该方程对应的椭圆系数bij(z)=bij2k+1(z,zn)=2bij2k(z,(4k+1)bij2k(z,2(4k+1)zn),k 0;bij2l+2(z,zn)=bij2k+1(z,2(4k+3)zn),k 0;bij2l1(z,zn)=bij2k(z,2(4k 1)zn),k 0;bij2l2(z,zn)=2bij2k1(z,(4k 3)bij2l1(z,2(4k 3)zn),k 0.由式(2.5),我们可得2k+=z Rn|z|1,zn=(1+4k),k Z,第 2 期包圆等：具有两个m-凸绝缘体和超导体内含物的电导问题的梯度估计5432k+1+=z Rn|z|1,zn=(1+4k),k Z.从而我们有2k+bnjzje w+2k+1+bnjzje w=0,k Z,C0(Q).因此Qbij(z)zje wzi=kQkbij(z)zje wzi=kQkbij(z)zje wzi+kk+bnjzje w=0.所以e w H1(Q)在Q中满足zj(bij(z)zje w)=0.我们在Q中引入一个新的方程zj(eBij(z)zju)=0,其中eBij(z)=limzQlz(0,(2k1)bij(z),z Qk,k 0;bij(0),z Q0;limzQlz(0,(2k+1)bij(z),z Qk,k 0.定义范数如下:GYs,p=sup0r1r1s(rQ|G|p)1p.注意到bij(z)C(Q0),bij(z)在区域Q中是分段C连续的,于是可以推出?bij(z)eBij(z)?Y1+,2 0是仅依赖于n,1,1的常数,和无关.3.主要结果及其证明定理1设u H1(r)是方程(1.1)的一个弱解.若边界条件f和g满足(2.1)-(2.3),则存在仅依赖于n,R0,fC2,gC2,的常数C,使得|u(x)|CuL(r)+|x|m,x (r2),n 2.(3.1)证不失一般性,本文假定r 1,以及式(2.3)中的a=1,即f(x)g(x)=|x|m+O(|x|m+),固定(x0,0)(r2),令=f(x0)g(x0)+.因为f和g满足(2.1)-(2.3),所以可以得到=f(x0)g(x0)+=|x|m+O(|x|m+)+C2|x0|m+.从而可得1C|x0|m+C|x0|m+,(3.2)544应用数学2024其中C仅依赖于DlC2,(l=1,2).考虑如下坐标变换y=(xx0),yn=xn.(3.3)令v(y)=u(x0+y,yn),aij(y)=aij(x0+y,yn),则在新的坐标系下,狭窄区域及其边界为e(r):=(y,yn)Rn?|y|r,g(x0+y)2 yn f(x0+y)+2,e+(r):=(y,yn)Rn?|y|r,yn=f(x0+y)+2,e(r):=(y,yn)Rn?|y|1m,则|x0|m2|x0|m+2|x0|m2,从而|x0|m+|x0|m2.令 r 12|x0|m212|x0|,由于|y|r,并且 r与 无关,从而有12|x0|x0+y|32|x0|,进而得到(12)m(|x0|m+)|x0+y|m+(32)m(|x0|m+).因此,|x0|m+|x0+y|m+.(3.7)于是由式(3.5)与(3.7)我们可推得式(3.6).2)若|x0|1m,则|x0|m+2,于是我们有|x0|m+.由于|y|r,且 r与无关,从而有y r r,进而|x0|m+|x0+y|m+.(3.8)因此由式(3.5)与(3.8)我们可推得式(3.6).注意到(yz)的特征值为 r(n 1重)和ynzn,由式(3.6)可知矩阵(yz)的所有特征值都被一致的常数控制,结合函数f和g的光滑性,可得t(bij(z)=t(yz)(aij(z)|det(yz)|(yz)t 1|2,Rn,?bij(z)?C(Q0)0和1 0是一致的常数.利用引理1,我们有wL(Q0(12)CwL(Q0).通过坐标变换(3.3)-(3.4)转化回关于u的方程,可得|u(x)|CuL(r)CuL(r)+|x|m,x (r2).于是式(3.1)得证.546应用数学2024参考文献：1 AMMARI H,GIRAOLO G,KANG H,et al.Spectral analysis of the Neumann-Poincar e operatorand characterization of the stress concentration in anti-plane elasticityJ.Arch.Ration.Mech.Anal.,2013,208(1):275-304.2 AMMARI H,DAVIES B,YU S.Close-to-touching acoustic subwavelength resonators:eigenfrequencyseparation and gradient blow-upJ.Multiscale Model.Simul.,2020,18(3):1299-1317.3 AMMARI H,KANG H,LEE H,et al.Optimal estimates for the electric field in two dimensionsJ.J.Math.Pures Appl.,2007,88(4):307-324.4 BAO E,LI Y Y,YIN B.Gradient 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estimate;Perfect;Insulated;m-convex