本篇文章,我们将研究$2m$阶振荡算子伴随着Dirichlet-Neumann混合边界条件的齐次化问题解的$H^{m}$和$L^{2}$收敛率.

具体地,令$\Omega$是${\Bbb R} ^{n}$中的有界Lipschitz区域.假设$\partial \Omega=\Gamma_{1}\bigcup\Gamma_{2}$,其中$\Gamma_{1}$和$\Gamma_{2}$分别是$\partial \Omega$的两个互不相交的闭子集.我们考虑如下问题

(1.1) $\begin{equation}\label{1.1} \left\{\begin{array}{ll}\displaystyle L_{\varepsilon}u_{\varepsilon}=(-1)^{m}\sum\limits_{|\alpha|=|\beta|=m}D^{\alpha}\left (A_{\alpha\beta}(x/\varepsilon)D^{\beta}u_{\varepsilon}\right)=f, &x\in\Omega, \\[2mm] \displaystyle Tr(D^{\gamma}u_{\varepsilon})=g_{\gamma}, &x\in\Gamma_{1}, ~0\leq |\gamma|\leq m-1, \\ \displaystyle N_{m-1-\gamma}(u_{\varepsilon})=h_{\gamma}, &x\in\Gamma_{2}, ~0\leq |\gamma|\leq m-1, \end{array} \right. \end{equation}$

这里的$\alpha, \beta, \gamma$是多重指标,其组成成分是$\alpha_{k}, \beta_{k}, \gamma_{k}$, $k=1, 2, \ldots, n$,并且

$|\alpha|=\sum\limits_{k=1}^{n}\alpha_{k}, ~~ D^{\alpha}=D^{\alpha_{1}}_{x_{1}}D^{\alpha_{2}}_{x_{2}} \cdots D^{\alpha_{n}}_{x_{n}}.$

为了书写方便,全文使用了求和约定.这里假设矩阵$A(y)=(A_{\alpha\beta}(y))$是实对称、有界可测的,并且满足

(1.2) $\begin{equation}\label{1.2} \sum\limits_{|\alpha|=|\beta|=m}A_{\alpha\beta}(y)\xi_{\alpha}\xi_{\beta}\geq \lambda \mid\xi\mid ^{2} , ~~\parallel A_{\alpha\beta}(y)\parallel_{L^{\infty}({\Bbb R} ^{n})}\leq \frac{1}{\lambda}, \end{equation}$

其中$y\in {\Bbb R} ^{n}, ~\xi=(\xi_{\alpha})_{|\alpha|=m}\in {\Bbb R} ^{n}, $这里的$\lambda> 0$;周期性条件

(1.3) $ \begin{equation}\label{1.3} A(y+Y)=A(y)~~\mbox{对于}~y\in {\Bbb R} ^{n}, ~~Y\in {\Bbb Z}^{n}. \end{equation}$

这里,还假设光滑性条件

(1.4) $\begin{equation}\label{1.4} f\in H^{-m}(\Omega), ~g_{\gamma}\in H^{m}(\partial\Omega), ~h_{\gamma}\in H^{-m/2}(\partial\Omega). \end{equation}$

伴随着问题(1.1)的齐次化问题是

(1.5) $\begin{equation}\label{1.6} \left\{\begin{array}{ll}\displaystyle L_{0}u_{0}=(-1)^{m}\sum\limits_{|\alpha|=|\beta|=m}D^{\alpha}\left(Q_{\alpha\beta}D^{\beta}u_{0}\right)=f, &x\in\Omega, \\ [2mm]\displaystyle Tr(D^{\gamma}u_{0})=g_{\gamma}, &x\in\Gamma_{1}, ~0\leq |\gamma|\leq m-1, \\ \displaystyle N_{m-1-\gamma}(u_{0})=h_{\gamma}, &x\in\Gamma_{2}, ~0\leq |\gamma|\leq m-1. \end{array} \right. \end{equation} $

这里的常系数矩阵$Q_{\alpha\beta}$,即所谓的齐次化矩阵,为

$Q_{\alpha\beta}=\sum\limits_{|\gamma|=m}\int_{Y}[A_{\alpha\beta}(y)+A_{\alpha\gamma}(y)D^{\gamma}\chi^{\gamma\beta}(y)]{\rm d}y, $

其中$Y=[0, 1)^{n}$.这里的$\chi=(\chi^{\gamma})$是算子$L_{\varepsilon}$的校正矩阵.

众所周知,解的收敛率研究是齐次化理论中的关键问题.二阶椭圆方程的齐次化问题结果较为丰富,读者可参阅文献[4-8, 18-19].相对来说,关于高阶椭圆方程的齐次化理论研究结果较少.最近,文献[1-3]得到了$L^{2}$空间下最优的$O(\varepsilon)$收敛率. 2017年, Suslina研究了具有两个参数的高阶椭圆方程组的齐次化问题(见文献[9-10]),其证明了在有界的$C^{2m}$区域上, Dirichlet和Neumann问题各自的$L^{2}$收敛率.最近,文献[11-12]研究了$2m$阶具有振荡系数的椭圆方程组的$H^{m-1}_{0}$收敛率. 2018年, Niu和Xu^[14]考虑了系数与时间相关的高阶抛物方程解的$L^{2}$收敛率.

据作者所知,关于高阶方程伴随着混合边界条件的齐次化问题,收敛率结果并不太多.本文将尝试着在此方向上进行突破.文中所用的方法,利用光滑算子巧妙避免了边界重叠项的估计,这对于处理混合边界问题既简单又直接.此工作也可视为将Suslina引入的光滑算子(见文献[9-10]),使用推广到高阶方程混合边界条件的情形.

以下是本文的主要结果.

定理1.1  令$\Omega$是${\Bbb R} ^{n}$中的有界Lipschitz区域.假设$u_{\varepsilon}\in H^{m}(\Omega)$, $u_{0}\in H^{m+1}(\Omega)$分别是混合边值问题(1.1)和(1.6)的两个弱解.当其满足条件(1.2)-(1.4)时,则有

$\bigg\| u_{\varepsilon}-u_{0}-\varepsilon^{m}\sum\limits_{|\gamma|=m}\chi^{\gamma} T_{\varepsilon}(D^{\gamma} \widetilde{u}_{0})\eta_{\varepsilon}\bigg\|_{H_{0}^{m}(\Omega)}\leq C\varepsilon^{1/2}\parallel u_{0}\parallel_{H^{m+1}(\Omega)}, $

其中$T_{\varepsilon}$是光滑算子, $\chi$是细胞问题的解, $\eta_{\varepsilon}$是截断函数.

定理1.2  在定理1.1的假设条件下,则有

$\parallel u_{\varepsilon}-u_{0}\parallel_{H_{0}^{m-1}(\Omega)}\leq C\varepsilon^{1/2}\parallel u_{0}\parallel_{H^{m+1}(\Omega)}. $

同时

$\parallel u_{\varepsilon}-u_{0}\parallel_{L^{2}(\Omega)}\leq C\varepsilon^{1/2}\parallel u_{0}\parallel_{H^{m+1}(\Omega)}. $

本文其他部分安排如下.第二部分包含了一些基本公式和有用的命题,它们对于得到收敛率结果至关重要.第三部分,证明得到了高阶方程混合边值问题解的$H_{0}^{m}(\Omega)$和$L^{2}$收敛率,证明过程中主要用到了光滑算子.第四部分,我们精炼了文中所得结果,同时讨论了下一步可能的研究方向.

我们首先介绍一些符号和定义.

令$B_{r}(x)=\{y\in {\Bbb R} ^{n}:\mid y-x\mid<r\}$表示球心在$x$点,半径是$r$的开球. $\Omega_{\varepsilon}=\{x\in \Omega: dist(x, \partial\Omega)\leq \varepsilon\}$, $\widetilde{\Omega}_{\varepsilon}=\{x\in {\Bbb R} ^{n}: dist(x, \partial\Omega)\leq \varepsilon\}$.令$H_{0}^{m}(\Omega;\Gamma_{1})$表示$\Omega$中的$H^{m}$函数族,并且其在$\Gamma_{1}$上等于0.由于区域$\Omega$是Lipschitz的,所以存在有界延拓算子$E:H^{m+1}(\Omega)\rightarrow H^{m+1}({\Bbb R} ^{n})$,用$\widetilde{u}_{0}=E(u_{0})$表示$u_{0}$的延拓,并且有$\parallel \widetilde{u}_{0}\parallel_{H^{m+1}({\Bbb R} ^{n})}\leq C\parallel u_{0}\parallel_{H^{m+1}(\Omega)}$.令$0\leq\eta_{\varepsilon}\leq 1$表示$C_{c}^{\infty}(\Omega)$中的截断函数,在$\Omega\backslash \Omega_{4\varepsilon}$上$\eta_{\varepsilon}=1$,并有$|\triangledown ^{k}\eta_{\varepsilon}|\leq C\varepsilon^{-k}$对于$1\leq k\leq m$成立,且有$supp (\eta_{\varepsilon})\subset \{x\in \Omega:dist(x, \partial \Omega)\geq 3\varepsilon\}=\Omega\backslash \Omega_{3\varepsilon}$.这里用$C$表示正常数,它在不同公式中可能取值不同.

这里的$ L_{0}$表示常系数算子,即齐次化算子.校正矩阵$\chi=(\chi^{\gamma}(y))$是以下细胞问题的解

$\begin{eqnarray*} \left\{\begin{array}{ll}\displaystyle\sum\limits_{|\alpha|=|\beta|=m} D^{\alpha}\left[A_{\alpha\beta}(y)D^{\beta}\chi^{\gamma}\right]=-\sum\limits_{|\alpha|=m}D^{\alpha}\left[A_{\alpha\gamma}(y)\right], &y\in Y, \\ \displaystyle\chi^{\gamma}(y+h)=\chi^{\gamma}(y), &y\in {\Bbb R} ^{n}, ~~h\in {\Bbb Y}^{n}, \\ \displaystyle\int_{Y}\chi^{\gamma}(y){\rm d}y=0. \end{array} \right. \end{eqnarray*}$

最近, Suslina引入了光滑算子(参见文献[9-10]),它主要用于建立椭圆或抛物算子的$L^{2}$收敛率.本工作试图将光滑算子的使用推广到了高阶方程的情形.

固定$\psi\in C^{\infty}_{0}(B_{1}(0))$,使其满足$\psi \geq 0$和$\int_{{\Bbb R} ^{n}}\psi {\rm d}x=1$.定义$L^{2}$中的算子$T_{\varepsilon}$,使其满足

$T_{\varepsilon}(u)(x)=u\ast\psi_{\varepsilon}=\int_{{\Bbb R} ^{n}}u(x-y)\psi_{\varepsilon}(y){\rm d}y, $

其中$\psi_{\varepsilon}(x)=\varepsilon^{-n}\psi(x/\varepsilon)$.

命题2.1  如果$u_{0}\in W^{m, p}({\Bbb R} ^{n})$对某些$1\leq p<\infty$成立,那么对于任意的多重指标$i$,当$|i|=m$时,都有估计式

$ \parallel u_{0}-T_{\varepsilon}( u_{0})\parallel_{L^{p}({\Bbb R} ^{n})}\leq C\varepsilon\parallel \bigtriangledown u_{0}\parallel_{L^{p}({\Bbb R} ^{n})}, $

$ \parallel T_{\varepsilon}(D^{i}u_{0})\parallel_{L^{2}(\Omega\backslash \Omega_{3\varepsilon})}\leq C\varepsilon^{-m}\parallel u_{0}\parallel_{L^{2}(\Omega\backslash \Omega_{2\varepsilon})} $

和

$\parallel T_{\varepsilon}(u_{0})\parallel_{L^{2}(\Omega_{\varepsilon})}\leq C\parallel u_{0}\parallel_{L^{2}(\widetilde{\Omega}_{2\varepsilon})}. $

证  这些估计式可以利用Parseval's定理和Hölder's不等式证明,参见文献[11].

命题2.2  令$F_{\alpha\beta}(y)\in L^{2}(Y)$表示周期函数,其中$Y=[0, 1)^{n}$.如果$ \int_{Y}F_{\alpha\beta}(y){\rm d}y=0$和$ \sum\limits_{|\alpha|=m}D^{\alpha}F_{\alpha\beta}(y)=0$对任意的$\beta$,当$|\beta|=m$成立.那么存在函数$\Phi_{\gamma\alpha\beta}\in H^{m}(Y)$,使得$ \sum\limits_{|\gamma|=m}D^{\gamma}\Phi_{\gamma\alpha\beta}=F_{\alpha\beta}$并且有$\Phi_{\gamma\alpha\beta}=-\Phi_{\alpha\gamma\beta}$.

证  这个命题往往被称为校正流性质.对于二阶线性算子的情形是众所周知的(如文献[18, Lemma 3.1]).上述命题由Pastukhova给出了证明(参见文献[21]).

注2.1  令

$F_{\alpha\beta}(y)=Q_{\alpha\beta}-A_{\alpha\beta}(y)-\sum\limits_{|\gamma|=m}A_{\alpha\gamma}(y)D^{\gamma}\chi^{\beta}(y). $

则周期函数$F_{\alpha\beta}(y)$满足性质$\int_Y F_{\alpha\beta}(y){\rm d}y=0$和$ \sum\limits_{|\alpha|=m}D^{\alpha}F_{\alpha\beta}=0$.从命题2.2,可知存在函数$\Phi_{\gamma\alpha\beta}$,满足$ \sum\limits_{|\gamma|=m}D^{\gamma}\Phi_{\gamma\alpha\beta} =F_{\alpha\beta}$和$\Phi_{\gamma\alpha\beta} =-\Phi_{\alpha\gamma\beta}$.

命题2.3  如果$u_{0}\in H^{m+1}({\Bbb R} ^{n})$,那么

$\parallel \triangledown^{m} u_{0} \parallel_{L^{2}(\widetilde{\Omega}_{\varepsilon})} \leq C\varepsilon^{1/2}\parallel u_{0}\parallel_{H^{m+1}({\Bbb R} ^{n})}. $

证  证明可在文献[20]或[23]中找到.

定义

(3.1) $M_{\varepsilon}=u_{\varepsilon}-u_{0}-\varepsilon^{m}\sum\limits_{|\gamma|=m}\chi^{\gamma} T_{\varepsilon}(D^{\gamma} \widetilde{u}_{0})\eta_{\varepsilon}. $

由$M_{\varepsilon}$的表达式不难看出,其是以下混合边值问题的弱解

(3.2) $\begin{equation}\label{3.1} \left\{\begin{array}{ll}\displaystyle L_{\varepsilon}M_{\varepsilon}=L_{0}u_{0}-L_{\varepsilon}u_{0}-L_{\varepsilon}\bigg(\varepsilon^{m}\sum\limits_{|\gamma|=m}\chi^{\gamma} T_{\varepsilon}(D^{\gamma} \widetilde{u}_{0})\eta_{\varepsilon}\bigg), &x\in\Omega, \\ \displaystyle Tr(D^{\gamma}M_{\varepsilon})=-\varepsilon^{m}\sum\limits_{|\gamma|=m}\chi^{\gamma} T_{\varepsilon}(D^{\gamma} \widetilde{u}_{0})\eta_{\varepsilon}, &x\in\Gamma_{1}, \\ \displaystyle N_{m-1-\gamma}(M_{\varepsilon})=N_{m-1-\gamma}(u_{\varepsilon })-N_{m-1-\gamma}(u_{0})\\ \quad\quad\quad\quad-N_{m-1-\gamma}\bigg(\varepsilon^{m}\sum\limits_{|\gamma|=m}\chi^{\gamma} T_{\varepsilon}(D^{\gamma} \widetilde{u}_{0})\eta_{\varepsilon}\bigg), &x\in\Gamma_{2}.\end{array} \right. \end{equation} $

这里我们利用了$u_{\varepsilon}$和$u_{0}$各自满足的方程(1.1)和(1.5).

事实上,对于任意的$\varphi\in H_{0}^{m}(\Omega;\Gamma_{1})$,都有

$\int_{\Omega}\sum\limits_{|\alpha|=|\beta|=m}D^{\alpha}\varphi A_{\alpha\beta} D^{\beta}u_{\varepsilon} {\rm d}x=\int_{\Omega}\sum\limits_{|\alpha|=|\beta|=m}D^{\alpha}\varphi Q_{\alpha\beta} D^{\beta}u_{0} {\rm d}x, $

由此可得

$\begin{eqnarray*}&&\int_{\Omega}\sum\limits_{|\alpha|=|\beta|=m}D^{\alpha}\varphi A_{\alpha\beta} D^{\beta}M_{\varepsilon} {\rm d}x\\&=&\sum\limits_{|\alpha|=|\beta|=m}\int_{\Omega}D^{\alpha}\varphi \bigg[Q_{\alpha\beta}D^{\beta} u_{0} -A_{\alpha\beta}D^{\beta}u_{0}-\varepsilon^{m}\sum\limits_{|\gamma|=m}A_{\alpha\beta}D^{\beta}\left(\chi^{\gamma} T_{\varepsilon}(D^{\gamma} \widetilde{u}_{0})\eta_{\varepsilon}\right)\bigg]{\rm d}x. \end{eqnarray*}$

经过简单计算,可知

$\begin{eqnarray*}&&Q_{\alpha\beta}D^{\beta} u_{0}-A_{\alpha\beta}D^{\beta}u_{0}-\varepsilon^{m} \sum\limits_{|\gamma|=m}A_{\alpha\beta}D^{\beta}\left(\chi^{\gamma} T_{\varepsilon}(D^{\gamma} \widetilde{u}_{0})\eta_{\varepsilon}\right)\\ &=&[Q_{\alpha\beta}(D^{\beta} u_{0}-T_{\varepsilon}(D^{\beta} \widetilde{u}_{0})\eta_{\varepsilon})-A_{\alpha\beta}(D^{\beta} u_{0}-T_{\varepsilon}(D^{\beta} \widetilde{u}_{0})\eta_{\varepsilon})]\\ && -\varepsilon^{m}\sum\limits_{|\gamma|=m} A_{\alpha\beta} \chi^{\gamma} T_{\varepsilon}(D^{\beta}D^{\gamma} \widetilde{u}_{0})\eta_{\varepsilon}\\ && -\varepsilon^{m}\sum\limits_{|\gamma|=m}\sum\limits_{i+j=\beta} A_{\alpha\beta} \chi^{\gamma} T_{\varepsilon}(D^{i}D^{\gamma} \widetilde{u}_{0})D^{j}\eta_{\varepsilon}\\ &&-\sum\limits_{|\gamma|=m}\sum\limits_{i+j=\beta}\varepsilon^{m-|i|}A_{\alpha\beta}D^{i}\chi T_{\varepsilon}(D^{j}D^{\gamma}u_{0})\eta_{\varepsilon} \\ &&-\sum\limits_{|\gamma|=m}\sum\limits_{i+j+k=\beta}\varepsilon^{m-|i|}A_{\alpha\beta}D^{i}\chi T_{\varepsilon}(D^{j}D^{\gamma}u_{0})D^{k}\eta_{\varepsilon}\\ &&+[Q_{\alpha\beta}-A_{\alpha\beta}-\sum\limits_{|\gamma|=m}A_{\alpha\beta}D^{\beta} \chi^{\gamma}]T_{\varepsilon}(D^{\gamma} \widetilde{u}_{0})\eta_{\varepsilon}. \end{eqnarray*} $

于是,就有

(3.3) $\begin{eqnarray}\label{3.00} &&\int_{\Omega}\sum\limits_{|\alpha|=|\beta|=m}D^{\alpha}\varphi A_{\alpha\beta} D^{\beta}M_{\varepsilon} {\rm d}x\\&=&\sum\limits_{|\alpha|=|\beta|=m}\int_{\Omega}D^{\alpha}\varphi[Q_{\alpha\beta}(D^{\beta} u_{0}-T_{\varepsilon}(D^{\beta} \widetilde{u}_{0})\eta_{\varepsilon})-A_{\alpha\beta}(D^{\beta} u_{0}-T_{\varepsilon}(D^{\beta} \widetilde{u}_{0})\eta_{\varepsilon})]{\rm d}x \nonumber\\&&-\sum\limits_{|\alpha|=|\beta|=|\gamma|=m}\int_{\Omega}\varepsilon^{m}D^{\alpha}\varphi A_{\alpha\beta} \chi^{\gamma} T_{\varepsilon}(D^{\beta}D^{\gamma} \widetilde{u}_{0})\eta_{\varepsilon}{\rm d}x \nonumber\\&&-\sum\limits_{|\alpha|=|\beta|=|\gamma|=m, i+j=\beta}\int_{\Omega}\varepsilon^{m}D^{\alpha}\varphi A_{\alpha\beta} \chi^{\gamma} T_{\varepsilon}(D^{i}D^{\gamma} \widetilde{u}_{0})D^{j}\eta_{\varepsilon}{\rm d}x \nonumber\\&&-\sum\limits_{|\alpha|=|\beta|=|\gamma|=m, i+j=\beta}\int_{\Omega}\varepsilon^{m-|i|}D^{\alpha}\varphi A_{\alpha\beta} D^{i}\chi^{\gamma} T_{\varepsilon}(D^{j}D^{\gamma} \widetilde{u}_{0})\eta_{\varepsilon}{\rm d}x \nonumber\\&&-\sum\limits_{|\alpha|=|\beta|=|\gamma|=m, i+j+k=\beta}\int_{\Omega}\varepsilon^{m-|i|}D^{\alpha}\varphi A_{\alpha\beta} D^{i}\chi^{\gamma} T_{\varepsilon}(D^{j}D^{\gamma} \widetilde{u}_{0})D^{k}\eta_{\varepsilon}{\rm d}x \nonumber\\&&+\sum\limits_{|\alpha|=|\beta|=m}\int_{\Omega}D^{\alpha}\varphi[Q_{\alpha\beta}-A_{\alpha\beta}-\sum\limits_{|\gamma|=m}A_{\alpha\beta}D^{\beta} \chi^{\gamma}]T_{\varepsilon}(D^{\gamma} \widetilde{u}_{0})\eta_{\varepsilon}{\rm d}x \nonumber\\&\doteq &I_{1}+I_{2}+I_{3}+I_{4}+I_{5}+I_{6}. \end{eqnarray}$

首先估计$I_{1}$.注意到

$\begin{array}{l}\displaystyle D^{\beta} u_{0}-T_{\varepsilon}(D^{\beta} \widetilde{u}_{0})\eta_{\varepsilon}\displaystyle=[D^{\beta}u_{0}-(D^{\beta} \widetilde{u}_{0})\eta_{\varepsilon}]+[D^{\beta} \widetilde{u}_{0}-T_{\varepsilon}(D^{\beta} \widetilde{u}_{0})]\eta_{\varepsilon}, \end{array}$

这里再次利用命题2.1和命题2.4,就有

(3.4) $\begin{eqnarray}\label{3.3} \mid I_{1}\mid &\displaystyle\leq &C\left(\parallel \triangledown^{m}u_{0}\parallel_{L^{2}(\Omega_{4\varepsilon})}+\parallel \triangledown^{m}\widetilde{u}_{0}-T_{\varepsilon}(\triangledown^{m}\widetilde{u}_{0})\parallel_{L^{2}(\Omega\backslash \Omega_{3\varepsilon})}\right)\parallel \triangledown^{m} \varphi\parallel_{L^{2}(\Omega)}\nonumber\\ &\leq& C \left(\varepsilon^{1/2}\parallel u_{0}\parallel_{H^{m+1}(\Omega)}+\varepsilon \parallel \triangledown^{m+1}\widetilde{u}_{0}\parallel_{L^{2}({\Bbb R} ^{n})}\right)\parallel \triangledown^{m} \varphi\parallel_{L^{2}(\Omega)}\nonumber\\ &\leq& C \varepsilon^{1/2}\parallel u_{0}\parallel_{H^{m+1}(\Omega)}\parallel \triangledown^{m} \varphi\parallel_{L^{2}(\Omega)}.\end{eqnarray}$

接下来估计$I_{2}$.由命题2.1可得

(3.5) $\begin{eqnarray}\label{3.5}\mid I_{2}\mid &\displaystyle\leq& C\varepsilon^{m}\parallel T_{\varepsilon}(\triangledown^{2m}\widetilde{u}_{0})\parallel_{L^{2}(\Omega\backslash \Omega_{3\varepsilon})}\parallel \triangledown^{m} \varphi\parallel_{L^{2}(\Omega)}\nonumber\\&\displaystyle\leq& C\varepsilon^{m}\varepsilon^{-m+1}\parallel T_{\varepsilon}(\triangledown^{m+1}\widetilde{u}_{0})\parallel_{L^{2}(\Omega\backslash \Omega_{2\varepsilon})}\parallel \triangledown^{m} \varphi\parallel_{L^{2}(\Omega)}\nonumber\\& \leq& C\varepsilon\parallel \triangledown^{m+1}u_{0}\parallel_{L^{2}(\Omega)}\parallel \triangledown^{m} \varphi\parallel_{L^{2}(\Omega)}\nonumber\\& \leq &C\varepsilon\parallel u_{0}\parallel_{H^{m+1}(\Omega)}\parallel \triangledown^{m} \varphi\parallel_{L^{2}(\Omega)}.\end{eqnarray}$

类似地,还可以得到

(3.6) $\begin{eqnarray}\displaystyle \mid I_{3}\mid &\displaystyle\leq &C\sum\limits_{1\leq h\leq m}\varepsilon^{m} \parallel T_{\varepsilon}(\triangledown^{2m-h}\widetilde{u}_{0})\triangledown^{h}\eta_{\varepsilon}\parallel_{L^{2}(\Omega)}\parallel \triangledown^{m} \varphi\parallel_{L^{2}(\Omega)} \nonumber\\ &\displaystyle\leq& C\varepsilon^{m}\varepsilon^{h-m} \parallel T_{\varepsilon}(\triangledown^{m}\widetilde{u}_{0})\triangledown^{h} \eta_{\varepsilon}\parallel_{L^{2}(\Omega)}\parallel \triangledown^{m} \varphi\parallel_{L^{2}(\Omega)} \nonumber\\ &\displaystyle\leq& C \parallel T_{\varepsilon}(\triangledown^{m}\widetilde{u}_{0})\parallel_{L^{2}(\Omega_{4\varepsilon}\backslash \Omega_{3\varepsilon})}\parallel \triangledown^{m} \varphi\parallel_{L^{2}(\Omega)} \nonumber\\ \displaystyle& \leq &C \parallel \triangledown^{m}\widetilde{u}_{0}\parallel_{L^{2}(\Omega_{5\varepsilon}\backslash \Omega_{3\varepsilon})}\parallel \triangledown^{m} \varphi\parallel_{L^{2}(\Omega)} \nonumber\\ \displaystyle& \leq &C\varepsilon^{1/2} \parallel u_{0}\parallel_{H^{m+1}(\Omega)}\parallel \triangledown^{m} \varphi\parallel_{L^{2}(\Omega)} \end{eqnarray}$

和

(3.7) $\begin{eqnarray}\mid I_{4}\mid &\displaystyle\leq &C\sum\limits_{1\leq h\leq m}\varepsilon^{m-h} \parallel \triangledown^{h}\chi T_{\varepsilon}(\triangledown^{2m-h}\widetilde{u}_{0})\eta_{\varepsilon}\parallel_{L^{2}(\Omega)}\parallel \triangledown^{m} \varphi\parallel_{L^{2}(\Omega)} \nonumber\\ &\displaystyle\leq& C\sum\limits_{1\leq h\leq m}\varepsilon^{m-h} \parallel T_{\varepsilon}(\triangledown^{2m-h}\widetilde{u}_{0})\parallel_{L^{2}(\Omega\backslash \Omega_{3\varepsilon})}\parallel \triangledown^{m} \varphi\parallel_{L^{2}(\Omega)}\nonumber\\ &\displaystyle\leq &C\varepsilon \parallel T_{\varepsilon}(\triangledown^{m+1}\widetilde{u}_{0})\parallel_{L^{2}(\Omega\backslash \Omega_{2\varepsilon})}\parallel \triangledown^{m} \varphi\parallel_{L^{2}(\Omega)} \nonumber\\ \displaystyle& \leq& C \varepsilon \parallel \triangledown^{m}\widetilde{u}_{0}\parallel_{L^{2}(\Omega\backslash \Omega_{\varepsilon})}\parallel \triangledown^{m} \varphi\parallel_{L^{2}(\Omega)}\nonumber\\ \displaystyle& \leq &C\varepsilon \parallel u_{0}\parallel_{H^{m+1}(\Omega)}\parallel \triangledown^{m} \varphi\parallel_{L^{2}(\Omega)}. \end{eqnarray} $

对于$I_{5}$,有如下估计

(3.8) $\begin{eqnarray} \mid I_{5}\mid &\displaystyle\leq &C\sum\limits_{k+h+l= m}\varepsilon^{m-l} \parallel \triangledown^{l}\chi T_{\varepsilon}(\triangledown^{m+h}\widetilde{u}_{0})\triangledown^{k}\eta_{\varepsilon}\parallel_{L^{2}(\Omega)}\parallel \triangledown^{m} \varphi\parallel_{L^{2}(\Omega)} \nonumber\\ &\displaystyle\leq& C\sum\limits_{1\leq h\leq m-2 }\varepsilon^{h} \parallel T_{\varepsilon}(\triangledown^{m+h}\widetilde{u}_{0})\parallel_{L^{2}(\Omega_{4\varepsilon}\backslash \Omega_{3\varepsilon})}\parallel \triangledown^{m} \varphi\parallel_{L^{2}(\Omega)} \nonumber\\ &\displaystyle\leq &C \parallel T_{\varepsilon}(\triangledown^{m}\widetilde{u}_{0})\parallel_{L^{2}(\Omega_{4\varepsilon}\backslash \Omega_{2\varepsilon})}\parallel \triangledown^{m} \varphi\parallel_{L^{2}(\Omega)}\nonumber\\ \displaystyle& \leq& C \parallel \triangledown^{m}\widetilde{u}_{0}\parallel_{L^{2}(\Omega_{5\varepsilon}\backslash \Omega_{2\varepsilon})}\parallel \triangledown^{m} \varphi\parallel_{L^{2}(\Omega)}\nonumber\\ \displaystyle& \leq &C\varepsilon^{1/2} \parallel u_{0}\parallel_{H^{m+1}(\Omega)}\parallel \triangledown^{m} \varphi\parallel_{L^{2}(\Omega)}. \end{eqnarray} $

最后还需估计$I_{6}$.令$F_{\alpha\beta}=Q_{\alpha\beta}-A_{\alpha\beta}-\sum\limits_{|\gamma|=m}A_{\alpha\beta}D^{\beta} \chi^{\gamma}$.注意到$F_{\alpha\beta}$是周期函数并且满足命题2.2的条件.利用注2.1,即存在周期函数$\Phi_{\gamma\alpha\beta}\in H^{m}(Y)$,满足性质$\Phi_{\gamma\alpha\beta}=-\Phi_{\alpha\gamma\beta}$和$ \sum\limits_{|\gamma|=m}D^{\gamma}\Phi_{\gamma\alpha\beta} =F_{\alpha\beta}$.

因此,利用散度定理,可得

$\begin{eqnarray*}I_{6}&\displaystyle=&\sum\limits_{|\alpha|=|\beta|=m}\int_{\Omega}D^{\alpha}\varphi[Q_{\alpha\beta}-A_{\alpha\beta}-\sum\limits_{|\gamma|=m}A_{\alpha\beta}D^{\beta} \chi^{\gamma}]T_{\varepsilon}(D^{\gamma} \widetilde{u}_{0})\eta_{\varepsilon}{\rm d}x \\ &\displaystyle =&\varepsilon^{m}\sum\limits_{|\alpha|=|\beta|=|\gamma|=m}\int_{\Omega}D^{\alpha}\varphi (D^{\gamma}\Phi_{\gamma\alpha\beta})T_{\varepsilon}(D^{\gamma} \widetilde{u}_{0})\eta_{\varepsilon}{\rm d}x \\ &\displaystyle =&\varepsilon^{m}\sum\limits_{|\alpha|=|\beta|=|\gamma|=m}(-1)^{m+1}\int_{\Omega}D^{\gamma}D^{\alpha}\varphi (\Phi_{\gamma\alpha\beta})T_{\varepsilon}(D^{\gamma} \widetilde{u}_{0})\eta_{\varepsilon}{\rm d}x \\&&-\varepsilon^{m}\sum\limits_{|\alpha|=|\beta|=|\gamma|=m, h+k=\gamma}\int_{\Omega}D^{h}D^{\alpha}\varphi (\Phi_{\gamma\alpha\beta})D^{k}[T_{\varepsilon}(D^{\gamma} \widetilde{u}_{0})\eta_{\varepsilon}]{\rm d}x \\&\displaystyle=&-\varepsilon^{m-|l|}\sum\limits_{|\alpha|=|\beta|=|\gamma|=m, l+k'=\gamma}(-1)^{|l|}\int_{\Omega}D^{\alpha}\varphi (D^{l}\Phi_{\gamma\alpha\beta})D^{k'}[T_{\varepsilon}(D^{\gamma} \widetilde{u}_{0})\eta_{\varepsilon}]{\rm d}x \\&\displaystyle=&-\varepsilon^{m-|l|}\sum\limits_{|\alpha|=|\beta|=|\gamma|=m, l+k_{1}+k_{2}=\gamma}(-1)^{|l|}\int_{\Omega}D^{\alpha}\varphi (D^{l}\Phi_{\gamma\alpha\beta})[T_{\varepsilon}(D^{k_{1}}D^{\gamma} \widetilde{u}_{0})D^{k_{2}}\eta_{\varepsilon}]{\rm d}x, \end{eqnarray*}$

这里,我们利用了函数$\Phi_{\gamma\alpha\beta}$的反对称性.

所以,利用命题2.1,可以得到

(3.9) $\begin{eqnarray}\mid I_{6}\mid &\displaystyle\leq &C\sum\limits_{0 \leq h\leq m-1}\varepsilon^{m-h} \parallel T_{\varepsilon}(\triangledown^{2m-h}\widetilde{u}_{0})\parallel_{L^{2}(\Omega\backslash \Omega_{3\varepsilon})}\parallel \triangledown^{m} \varphi\parallel_{L^{2}(\Omega)}\nonumber\\ &&+C\sum\limits_{0 \leq h\leq m-1}\varepsilon^{h} \parallel T_{\varepsilon}(\triangledown^{m+h}\widetilde{u}_{0})\parallel_{L^{2}(\Omega_{4\varepsilon}\backslash \Omega_{3\varepsilon})}\parallel \triangledown^{m} \varphi\parallel_{L^{2}(\Omega)}\nonumber\\ &\displaystyle\leq& C \parallel T_{\varepsilon}(\triangledown^{m}\widetilde{u}_{0})\parallel_{L^{2}(\Omega \backslash \Omega_{3\varepsilon})}\parallel \triangledown^{m} \varphi\parallel_{L^{2}(\Omega)}\nonumber\\ \displaystyle& \leq &C \parallel \triangledown^{m}\widetilde{u}_{0}\parallel_{L^{2}(\Omega\backslash \Omega_{2\varepsilon})}\parallel \triangledown^{m} \varphi\parallel_{L^{2}(\Omega)} \nonumber\\ \displaystyle& \leq &C\varepsilon^{1/2} \parallel u_{0}\parallel_{H^{m+1}(\Omega)}\parallel \triangledown^{m} \varphi\parallel_{L^{2}(\Omega)}. \end{eqnarray} $

综合估计式(3.3)-(3.9),就有

$\bigg| \int_{\Omega}\sum\limits_{|\alpha|=|\beta|=m}D^{\alpha}\varphi A_{\alpha\beta} D^{\beta}M_{\varepsilon} {\rm d}x \bigg|\leq C\varepsilon^{1/2}\parallel u_{0}\parallel_{H^{m+1}(\Omega)}\parallel \triangledown^{m}\varphi \parallel_{L^{2}(\Omega)}.$

取检验函数$\varphi=M_{\varepsilon}$,并且利用双线性形式的强制性条件,就可得到所要结论.这就完成了定理1.1的证明.

同时,注意到

$\begin{eqnarray*}&&\bigg\| \varepsilon^{m}\sum\limits_{|\gamma|=m}\chi^{\gamma} T_{\varepsilon}(D^{\gamma} \widetilde{u}_{0})\eta_{\varepsilon}\bigg\|_{H_{0}^{m-1}(\Omega)} \\&\leq &C\varepsilon^{m}\sum\limits_{|\gamma|=m}\sum\limits_{|k+i+j|=m-1}\parallel \varepsilon^{-k}(D^{k}\chi^{\gamma})(T_{\varepsilon}(D^{i}D^{\gamma} \widetilde{u}_{0}))(D^{j}\eta_{\varepsilon})\parallel_{L^{2}(\Omega)}\\&\leq &C\sum\limits_{0\leq h\leq m-1}\varepsilon^{h+1}\parallel T_{\varepsilon}(\triangledown^{m+h}\widetilde{u}_{0})\parallel_{L^{2}(\Omega)} \\ &\leq& C\varepsilon^{2} \parallel T_{\varepsilon}(\triangledown^{m+1}\widetilde{u}_{0})\parallel_{L^{2}({\Bbb R} ^{n})} \\ &\leq& C\varepsilon^{2} \parallel \triangledown^{m+1}\widetilde{u}_{0}\parallel_{L^{2}({\Bbb R} ^{n})} \\ & \leq& C\varepsilon^{2} \parallel u_{0}\parallel_{H^{m+1}(\Omega)}. \end{eqnarray*}$

这里,再结合定理1.1和Minkowski不等式,即证得定理1.2.

本篇文章研究了高阶偏微分方程Dirichlet-Neumann混合边界条件的齐次化问题解的收敛率.这里主要利用了光滑算子,这对于处理边界重叠项更加简单直接.具体地,我们利用齐次化手段和能量估计,对$2m$阶椭圆方程进行了收敛率定量估计.作为结果,本文得到了$H_{0}^{m}$和$L^{2}$收敛率.本文将经典的二阶椭圆方程的收敛结果推广到了高阶方程混合边界条件的情形.

据作者所知,前人的工作已经证明了经典齐次化问题在$L^{2}$空间中的最优收敛率应该是$O(\varepsilon)$.因此,本文所得到的$L^{2}$中的收敛率应该不是最优的.事实上,作者还希望得到对任意的$1\leq p<\infty$, $W^{m-1, p}$空间的收敛率,类似的结果对于高阶方程混合边界条件的齐次化问题还是未知的.这些都是很有趣的问题,将会在后续工作中深入研究.

虽然目前的工作还是纯理论的,但是希望本文的结果能致力于更好地理解高阶方程混合边界条件的齐次化问题.