对于各向同性,均匀介质的平面弹性问题,当Lamé常数$\lambda\rightarrow \infty$时,弹性材料是几乎不可压的,此时的协调有限元会产生闭锁现象^[1-5],即有限元解往往不能很好的收敛到原问题的真解,或者不能得到最优的收敛阶.产生此现象的原因是误差估计的系数中含有$\lambda$,不可压条件下误差估计会趋向于无穷.为了克服闭锁现象,需要构造满足特殊条件的单元,使得当单元剖分足够细时,对任意$\lambda$仍然能够一致收敛.

对于此问题的研究与讨论,之前有一些工作.文献[2-3, 6-8]讨论了纯应力边界条件的弹性问题,构造了满足第二类离散Korn不等式的有限元空间,并借助于混合有限元方法分析其收敛性.其中文献[2, 6]构造了三角形单元,文献[3, 7]构造了矩形单元,文献[8]构造了NRQ$_1$元.但是对于纯位移问题, Korn不等式不是构造Locking-free单元的必须条件.利用插值算子的散度满足散度的$L^2$投影,文献[1]通过三角非协调C-R元,对纯位移问题通过扩展有限元空间的分析方法得到误差估计,文献[1, 3]同时也讨论了纯应力问题的Locking-free单元,并对纯应力问题证明了Korn不等式.利用降阶积分的方法^[9]构造了矩形高次元和一次三角形元.文献[10]构造了一次四边形单元,在误差分析中采用将大单元分割成小单元的棋盘模式得到最优收敛阶.文献[11]给出了构造Locking-free单元的新途径,并且按照文中的方法构造了一个矩形元.文献[12]在^[11]基础上构造了一个最低阶的矩形Locking-free单元.文献[13]在三维空间讨论了弹性Locking-free四面体单元.文献[14]构造了形函数空间具有高阶项的矩形各向异性单元.文献[15]通过引入补充空间和新技巧构造了非协调具有各向异性的矩形元.文献[16-17]构造了位移散度属于$P_1$的多项式空间,且能量模具有二阶收敛, $L^2$模具有三阶收敛的三角形和矩形Locking-free单元.文献[18]利用双参数方法构造四面体单元并且通过后处理算子得到四边形单元的超收敛结果.

本文以弹性纯位移变分问题为基础构造了三维的三棱柱单元并且证明了单元的适定性,得到能量模具有一阶收敛性, $L^2$模具有二阶收敛性.

本文中, $X$是剖分单元$K$, $\hat{K}$或整个区域$\Omega$. $\|\cdot\|_{0, \, X}$定义$L^2(X)$空间中的范数, $H^k(X)$空间中函数的$k$阶导数属于$L^2(X)$,其上的范数和半范数定义为$\parallel \cdot \parallel_{k, \, X}$和$|\cdot|_{k, \, X}$,即

$ \parallel u \parallel_{k, \, X}= \bigg(\sum\limits_{0\leq|\alpha|\leq k}\|D^{\alpha}u\|^{2}_{0, \, X} \bigg)^{\frac{1}{2}}, \quad | u |_{k, \, X}= \bigg(\sum\limits_{|\alpha|=k}\|D^{\alpha}u\|^{2}_{0, \, X}\bigg)^{\frac{1}{2}}. $

$D^{\alpha} u$是函数$u$的弱导数. $H^k_0(X)$定义函数$v\in H^k(X)$且$v\mid_{\partial X}=0$.在三维空间中定义三维向量$\vec{u}\in(L^2(X))^3$,其中每一个分量均是$L^2(X)$中的元素.类似的$(H^k(X))^3$中每一个分量均属于$H^k(X)$. $P_{k}(X)$是$X$上的次数不超过$k$次的多项式空间. $\vec{u}$表示三维向量函数, $\vec{n}$表示单元的外法向分量. $\operatorname{div} \vec{u}$表示向量的散度,且$\operatorname{div} \vec{u}=\frac{\partial\vec{u}}{\partial x}+\frac{\partial\vec{u}}{\partial y}+\frac{\partial\vec{u}}{\partial z}.$

本文讨论的基本问题是均匀介质各向同性材料的纯位移平面弹性问题.令$\Omega\subset R^3$是凸区域,位移$\vec{u}=(u_1, \, u_2, \, u_3)\in(H^2(\Omega))^3$,满足

(2.1) $\begin{equation}\label{2.1} \left\{\begin{array}{ll} -\mu\vartriangle\vec{u}-(\mu+\lambda){\rm grad}(\operatorname{div}\vec{u}) = \vec{f}, & {\rm in} \ \Omega, \\ \vec{u}=0, & {\rm on} \ \partial\Omega. \end{array}\right.\end{equation}$

其中$\vec{f}\in(L^2(\Omega))^3, \, (\mu, \, \lambda)\in\, [\mu_1, \mu_2]\times(0, \, +\infty)$是Lamé常数,且$0 < \mu_1 < \mu_2$.

对应(2.1)变分问题:求$\vec{u}\in (H_0^1(\Omega))^3$满足

(2.2) $\begin{equation}\label{2.2} a(\vec{u}, \, \vec{v})=(\vec{f}, \, \vec{v}), \quad\forall \vec{v}\in\, (H_0^1(\Omega))^3.\end{equation}$

其中

(2.3) $ \begin{eqnarray}\label{2.3} a(\vec{u}, \, \vec{v})&=&\int_{\Omega}[\mu\nabla\vec{u}\, :\, \nabla\vec{v}+(\mu+\lambda)\operatorname{div}\vec{u}\cdot \operatorname{div}\vec{v}]{\rm d}x, \\ (\vec{f}, \, \vec{v})&=&\int_{\Omega}\vec{f}\cdot\vec{v}{\rm d}x. \end{eqnarray}$

显然由Korn不等式可知任意$\vec{v}\in(H_0^1(\Omega))^3$有

(2.4) $\begin{equation}\label{2.4} a(\vec{v}, \, \vec{v})=\mu|\vec{v}|^{2}_{1, \, \Omega}+(\mu+\lambda)\|\operatorname{div} \vec{v} \|^{2}_{0, \, \Omega}\geq \alpha\|\vec{v}\|^{2}_{1, \, \Omega}, \end{equation}$

由Lax-milgram定理可以得到上述变分问题有唯一解.

令$\Gamma_h$是三维空间$\Omega$的拟一致三棱柱剖分,且$\bar{\Omega}=\bigcup\limits_{K\in \Gamma_h}K$.三棱柱的上下两个平行的面与坐标面$z=0$平行.其它三个侧面与$z$轴平行.

令$a_i(x_i, \, y_i, \, z_i)(K)\, (i=1, \cdots, 6)$和$F_j(K)\, (j=1, \cdots, 5)$分别是单元$K$的顶点和外表面.令$\hat{K}$ (图 1)是参考单元,其坐标轴设为$\xi, \, \eta, \, \zeta$分别对应坐标轴$x, \, y, \, z$. $\hat{F}_j\, (j=1, \, \cdots, \, 5), \, \hat{a}_i\, (i=1, \, \cdots, \, 6)$分别是参考单元的顶点和面,令$h=\max\limits_{K\in \Gamma_h}h_K$.参考单元顶点坐标$\hat{a}_i$ (图 1).

在参考单元$\hat{K}$上定义多项式空间如下

(3.1) $ \begin{eqnarray}\label{3.2}\vec{\hat{P}}\triangleq \{(P_1(\hat{K}))^3\oplus(\zeta^2-\xi\zeta)\vec{p}_0\oplus(\eta^2-\xi\eta)\vec{q}_0\oplus(\xi^2-\xi\zeta)\vec{r}_0;\, \, \vec{p}_0, \, \vec{q}_0, \, \vec{r}_0\in \mathbb{R}^3\}, \end{eqnarray}$

其中$(P_1(\hat{K}))^3$是$R^3$中各个分量次数不超过一次的多项式空间,形函数空间定义如下

(3.2) $\begin{equation}\label{3.3}\hat{P}(\hat{K})\triangleq \{\vec{\hat{v}}|\vec{\hat{v}}\in\vec{\hat{P}}, \, \operatorname{div}\vec{\hat{v}}\in P_0(\hat{K})\}, \end{equation}$

则有dim$\hat{P}(\hat{K})=3\times 4+9-3=18$.

定理3.1   参考单元上的自由度定义如下

(3.3) $\left\{\begin{array}{l}{\text { (i) } \frac{1}{\left|\hat{F}_{i}\right|} \int_{\hat{F}_{i}} \overrightarrow{\hat{v}} \mathrm{d} \hat{f}, \quad i=1, \cdots, 5} \\ {\text { (ii) } \frac{1}{|\hat{K}|} \int_{\tilde{K}} \vec{v} \mathrm{d} \hat{x}}\end{array}\right.$

其中$\vec{\hat{v}}\in \hat{P}(\hat{K})$由(3.3)式唯一决定.

证   三棱柱的五个外表面的方程分别为$\hat{F}_1:\zeta=1, \, \hat{F}_2:\zeta=0, \, \hat{F}_3:\xi=0, \, \hat{F}_4:\eta=0, \, \hat{F}_5:1-\xi-\eta=0.$设

(3.4) $\begin{eqnarray}\label{3.5.1} \vec{\hat{v}}&= &\left(\begin{array}{ccc} \alpha_0+\alpha_1\xi+\alpha_2\eta+\alpha_3\zeta\\ \beta_0+\beta_1\xi+\beta_2\eta+\beta_3\zeta\\ \gamma_0+\gamma_1\xi+\gamma_2\eta+\gamma_3\zeta\\ \end{array}\right)+(\zeta^2-\xi\zeta)\left(\begin{array}{ccc}p_1 \\p_2\\p_3\\ \end{array}\right)\\& &+(\eta^2-\xi\eta)\left(\begin{array}{ccc}q_1 \\q_2\\q_3\\ \end{array}\right)+(\xi^2-\xi\zeta)\left(\begin{array}{ccc}r_1 \\r_2\\r_3\\ \end{array}\right). \end{eqnarray}$

由(3.3)式的第(ⅰ)个自由度和形函数空间(3.4)定义,可以得到下面的方程

$ \begin{eqnarray*} \int_{\hat{F}_1}\hat{v}_1{\rm d}\hat{f}=\int_{0}^{1}{\rm d}\xi\int_{0}^{1-\xi}\hat{v}_1{\rm d}\eta=\frac{1}{2}\alpha_0 +\frac{1}{6}\alpha_1+\frac{1}{6}\alpha_2+\frac{1}{2}\alpha_3+\frac{1}{3}p_1+\frac{1}{24}q_1-\frac{1}{12}r_1, \end{eqnarray*}$

同理在另外四个面上积分得下面四个方程

(3.5) $ \begin{eqnarray}\label{3.6} \int_{\hat{F}_2}\hat{v}_1{\rm d}\hat{f}&=&\frac{1}{2}\alpha_0 +\frac{1}{6}\alpha_1+\frac{1}{6}\alpha_2+\frac{1}{24}q_1+\frac{1}{12}r_1, \\ \int_{\hat{F}_3}\hat{v}_1{\rm d}\hat{f}&=&\alpha_0 +\frac{1}{2}\alpha_2+\frac{1}{2}\alpha_3+\frac{1}{3}p_1+\frac{1}{3}q_1, \\ \int_{\hat{F}_4}\hat{v}_1{\rm d}\hat{f}&=&\alpha_0 +\frac{1}{2}\alpha_1+\frac{1}{2}\alpha_3+\frac{1}{12}p_1+\frac{1}{12}r_1, \\ \int_{\hat{F}_5}\hat{v}_1{\rm d}\hat{f}&=&\sqrt{2} \bigg (\alpha_0+\frac{1}{2}\alpha_1+\frac{1}{2}\alpha_2+\frac{1}{2}\alpha_3+\frac{1}{12}p_1+\frac{1}{6}q_1+\frac{1}{12}r_1\bigg). \end{eqnarray}$

同理,对$\hat{v}_2, \, \hat{v_3}$采用同样的方法对三棱柱的各个面分别计算,可以得到另外两组类似于(3.5)式的方程组.再根据(3.3)式的第(ⅱ)个自由度,有

(3.6) $\begin{eqnarray}\label{3.7} \int_{\hat{K}_1}\hat{v}_1{\rm d}\hat{V}&=&\int_{0}^{1} {\rm d}\zeta\int_{0}^{1}{\rm d}\xi\int_{0}^{1-\xi}\hat{v}_1{\rm d}\eta=\frac{1}{2}\alpha_0 +\frac{1}{6}\alpha_1+\frac{1}{6}\alpha_2+\frac{1}{4}\alpha_3+\frac{1}{12}p_1+\frac{1}{24}q_1, \\ \int_{\hat{K}_1}\hat{v}_2{\rm d}\hat{V}&=&\frac{1}{2}\beta_0 +\frac{1}{6}\beta_1+\frac{1}{6}\beta_2+\frac{1}{4}\beta_3+\frac{1}{12}p_2+\frac{1}{24}q_2, \\ \int_{\hat{K}_1}\hat{v}_3{\rm d}\hat{V}&=&\frac{1}{2}\gamma_0 +\frac{1}{6}\gamma_1+\frac{1}{6}\gamma_2+\frac{1}{4}\gamma_3+\frac{1}{12}p_3+\frac{1}{24}q_3. \end{eqnarray}$

对(3.4)式定义的多项式求偏导数,得

(3.7) $\begin{eqnarray}\label{3.8}\frac{\partial \hat{v}_1}{\partial \xi}&=&\alpha_1-\zeta p_1-\eta q_1+(2\xi-\zeta)r_1, \\\frac{\partial \hat{v}_2}{\partial \eta}&=&\beta_2+(2\eta-\xi)q_2, \\\frac{\partial \hat{v}_3}{\partial \zeta}&=&\gamma_3+(2\zeta-\xi)p_3-\xi r_3. \end{eqnarray}$

由(3.2)式定义的形函数空间的约束条件div$ \vec{\hat{v}}\in P_0(\hat{K})$,得到下面三个方程

(3.8) $\begin{eqnarray}\label{3.9}&&2r_1-q_2-p_3-r_3=0, \\&&-q_1+2q_2=0, \\&&-p_1-r_1+2p_3=0. \end{eqnarray}$

计算由(3.5)式, (3.6)和(3.8)式构成的21个方程组成的方程组,可以得到矩阵的阶数也是21,即方程组有唯一解.也即形函数空间(3.2)可以由(3.3)式定义的自由度唯一确定.

定义仿射变换$F_K: \, \hat{K}\rightarrow K$,且

(3.9) $\begin{equation}\label{3.10}X=B\hat{X}+X_1, \end{equation}$

其中$\hat{a}_i\rightarrow a_i$, $X_1$指顶点a₁的坐标.计算可得

$ B=\left(\begin{array}{ccc}x_2-x_1 & x_3-x_1 & 0\\y_2-y_1 & y_3-y_1 & 0\\0 & 0 & z_4-z_1\\ \end{array}\right).$

对向量值函数采用Piola变换$\vec{v}=B\vec{\hat{v}}\cdot F^{-1}_K$,定义梯度算子

(3.10) $\begin{equation}\hat{\nabla}=\bigg(\displaystyle\frac{\partial}{\partial \xi}, \, \displaystyle\frac{\partial}{\partial \eta}, \, \displaystyle\frac{\partial}{\partial \zeta}\bigg)'=\hat{\rm grad}, \, \quad\nabla=\bigg(\displaystyle\frac{\partial}{\partial x}, \, \displaystyle\frac{\partial}{\partial y}, \displaystyle\frac{\partial}{\partial z}\bigg)'={\rm grad}, \label{3.11}\end{equation}$

由梯度算子的定义有$\hat{\nabla}=B^T\nabla$,因为

$\begin{eqnarray*}\hat{\nabla}=\left(\begin{array}{ccc}\displaystyle\frac{\partial}{\partial \xi} \\\displaystyle\frac{\partial}{\partial \eta} \\ \displaystyle\frac{\partial}{\partial \zeta} \end{array}\right)= \left(\begin{array}{ccc}\displaystyle\frac{\partial\displaystyle}{\partial x}\frac{\partial x}{\partial \xi}+\frac{\partial}{\partial y}\frac{\partial y}{\partial \xi}+\frac{\partial}{\partial z}\frac{\partial z}{\partial \xi} \\ \displaystyle \displaystyle\frac{\partial}{\partial x}\frac{\partial x}{\partial \eta}+\frac{\partial}{\partial y}\frac{\partial y}{\partial \eta}+\frac{\partial} {\partial z}\frac{\partial z}{\partial \eta} \\ \displaystyle\frac{\partial}{\partial x}\frac{\partial x}{\partial \zeta}+\frac{\partial}{\partial y}\frac{\partial y}{\partial \zeta}+\frac{\partial}{\partial z}\frac{\partial z}{\partial \zeta} \end{array}\right)=\left(\begin{array}{ccc}\displaystyle\frac{\partial x}{\partial \xi} &\displaystyle\frac{\partial y}{\partial \xi} & \displaystyle\frac{\partial z}{\partial \xi} \\ \displaystyle\frac{\partial x}{\partial \eta} & \displaystyle\frac{\partial y} {\partial \eta} & \displaystyle\frac{\partial z}{\partial \eta} \\ \displaystyle\frac{\partial x}{\partial \zeta} & \displaystyle\frac{\partial y}{\partial \zeta} & \displaystyle\frac{\partial z}{\partial \zeta} \end{array}\right) \left(\begin{array}{ccc}\displaystyle\frac{\partial}{\partial x} \\\displaystyle\displaystyle \frac{\partial}{\partial y} \\ \displaystyle\frac{\partial}{\partial z} \end{array}\right) =B^T\nabla.\end{eqnarray*}$

定义一般单元$K$上的形函数空间如下

(3.11) $\begin{equation}\label{3.12}P(K)=\{\vec{v}=B\vec{\hat{v}}\cdot F^{-1}_K, \quad \forall \vec{\hat{v}}\in \hat{P}(\hat{K})\}, \end{equation}$

定义插值算子$\Pi_K:\, (H^1(K))^3\rightarrow P(K)$和$\hat{\Pi}_{\hat{K}}:\, (H^1(\hat{K}))^3\rightarrow \hat{P}(\hat{K})$,且

(3.12) $\begin{equation}\label{3.13}\Pi_K \vec{v}=B\hat{\Pi}_{\hat{K}} \vec{\hat{v}}\cdot F^{-1}_K, \end{equation}$

显然插值算子是仿射等价的.

$\widehat{\Pi_K \vec{v}}=B^{-1}\Pi_K \vec{v}=\hat{\Pi} \vec{\hat{v}}|_{K}.$

对于任意$v\in(H^1(\Omega))^3$,定义$\Pi_h:\, \Pi_h\vec{v}|_K=\Pi_K\vec{v}$,定义一般单元上的形函数空间

(3.13) $\begin{equation}\label{3.14} V_h=\bigg\{\vec{v}_h|_K\in P(K), \, \int_{F_i}[\vec{v}_h]{\rm d}f=0, \forall F_i\subset \partial K\bigg\}, \end{equation} $

由有限元空间定义可知$v_h\nsubseteq (H^1_0(\Omega))^3$,因此这里构造的单元是非协调元.

引理3.1   插值算子$\hat{\Pi}$保持一次多项式空间不变,即

(3.14) $\begin{equation}\label{3.15} \hat{\Pi}\vec{\hat{p}}=\vec{\hat{p}}, \quad \forall \vec{\hat{p}}\in(P_1(\hat{K}))^3. \end{equation}$

引理3.2   对于任给$\vec{v}\in (H^1(\Omega))^3$,有$\operatorname{div} \vec{v}=\hat{\operatorname{div}}\vec{\hat{v}}$成立.

证   由$\hat{\nabla}=B^T\nabla, $得

$\hat{\operatorname{div}}\vec{\hat{v}}=\hat{\nabla}\cdot \vec{\hat{v}}=\hat{\nabla}^T\vec{\hat{v}}=(B^T\nabla)^T \vec{\hat{v}}=\nabla^TB\vec{\hat{v}}=\nabla^T\vec{v}=\operatorname{div}\vec{v}.$

证毕.

接下来我们进行误差估计,方程(2.2)对应的离散有限元格式为:求$\vec{u}_h\in V_h$

(4.1) $\begin{equation}\label{4.1} a_h(\vec{u}_h, \, \vec{v}_h)=(\vec{f}, \, \vec{v}_h), \quad\forall \vec{v}_h\in\, V_h, \end{equation}$

其中

(4.2) $\begin{eqnarray}\label{4.2}a_h(\vec{u}_h, \, \vec{v}_h)&=&\sum\limits_{K\in \Gamma_h}\int_{K}[\mu\nabla\vec{u}_h\, :\, \nabla\vec{v}_h+(\mu+\lambda)\operatorname{div}\vec{u}_h\cdot \operatorname{div}\vec{v}_h]{\rm d}x, \\(\vec{f}, \, \vec{v}_h)&=&\int_{\Omega}\vec{f}\cdot\vec{v}_h{\rm d}x. \end{eqnarray}$

定义有限元空间$V_h$上的范数

(4.3) $\begin{equation}\label{4.3}\|\vec{v}_h\|_h=(\mu|\vec{v}_h|_1^2+(\mu+\lambda)\|\operatorname{div} \vec{v}_h\|_0^2)^{\frac{1}{2}}=(a_h(\vec{v}_h, \, \vec{v}_h))^{\frac{1}{2}}, \end{equation}$

显然当$\|\vec{v}_h\|=0$时得到$\vec{v}_h=C$ (常数).再由有限元空间定义,在$\Omega$边界上有$\int_{\partial\Omega}[\vec{v}_h]{\rm d}f=0$,则可知$\vec{v}_h|_{\Omega}\equiv 0$.

引理4.1   设$L^2$投影算子$\Pi_{P_0}:\, L^2(K)\rightarrow P_0$满足$\int_K\Pi_{P_0}w{\rm d}x=\int_K w{\rm d}x$, $\forall w\in L^2(K)$,则有

(4.4) $ \begin{equation}\label{4.4} \Pi_{P_0}\operatorname{div}\vec{v}=\operatorname{div}\Pi_K\vec{v}, \quad \forall\vec{v}\in H^1(K). \end{equation} $

证   由单元$K$上自由度定义和Green公式,得

$\int_{K} \operatorname{div} \Pi_{K} \vec{v} \mathrm{d} x=\int_{\partial K} \Pi_{K} \vec{v} \cdot \vec{n} \mathrm{d} F=\int_{\partial K} \vec{v} \cdot \vec{n} \mathrm{d} F=\int_{K} \operatorname{div} \vec{v} \mathrm{d} x$

证毕.

引理4.2   定义$P_0\vec{v}=\frac{1}{|F_i|}\int_{F_i}\vec{v}{\rm d}F$,并且设$ L(\frac{\partial\vec{u}}{\partial\vec{n}})$是$\frac{\partial\vec{u}}{\partial\vec{n}}$的分片常数插值.则

(4.5) $ \begin{equation}\label{4.5} \sum\limits_{K\in\Gamma_h}\sum\limits_{i=1}^{5}\int_{F_i}L\bigg(\frac{\partial\vec{u}}{\partial\vec{n}}\bigg)(\vec{v}-P_0\vec{v}){\rm d}F=0. \end{equation} $

证

$\sum\limits_{K\in\Gamma_h}\sum\limits_{i=1}^{5}\int_{F_i}L\bigg(\frac{\partial\vec{u}}{\partial\vec{n}}\bigg)(\vec{v}-P_0\vec{v}){\rm d}F= \sum\limits_{K\in\Gamma_h}\sum\limits_{i=1}^{5}L\bigg(\frac{\partial\vec{u}}{\partial\vec{n}}\bigg)\int_{F_i}(\vec{v}-P_0\vec{v}){\rm d}F=0. $

证毕.

定理4.1   令$\Gamma_h$是$\Omega$的正则拟一致剖分,设$\vec{f}\in (L^2(\Omega))^3$,令$\vec{u}\in(H_0^1(\Omega))^3\cap (H^2(\Omega))^3$是连续问题$(2.1)$的解,且$\vec{u}_h\in V_h$是离散问题$(4.1)$的解,则有下述误差估计成立

(4.6) $ \begin{equation}\label{4.6} \|\vec{u}-\vec{u}_h\|_h\leq ch(|\vec{u}|_{2, \, \Omega}+\lambda|\operatorname{div} \vec{u}|_{1, \, \Omega}). \end{equation}$

证   由Strong引理

(4.7) $\begin{equation}\label{4.7} \|\vec{u}-\vec{u}_h\|_h\leq \inf\limits_{\vec{v}\in V_h}\|\vec{u}-\vec{v}_h\|_h+\sup\limits_{\vec{v}\in V_h\setminus \{0\}}\frac{|a_h(\vec{u}, \vec{v}_h)-\int_{\Omega}\vec{f}\cdot \vec{v}_h{\rm d}x|}{\|\vec{v}_h\|_h}. \end{equation} $

首先讨论逼近误差,由于插值对一次多项式精确成立,所以有

(4.8) $\begin{equation}\label{4.8} |\vec{u}-\Pi_h\vec{u}|_{1, \, K}\leq ch|\vec{u}|_{2, \, K}, \end{equation}$

再由引理4.1得

(4.9) $\begin{equation}\label{4.9} \|\operatorname{div} (\vec{u}-\Pi_K\vec{u})\|_{0, \, K}=\|\operatorname{div} \vec{u}-\Pi_{P_0}\operatorname{div}\vec{u}\|_{0, \, K}\leq ch |\operatorname{div} \vec{u}|_{1, \, K}. \end{equation} $

由(4.7)-(4.9)式得逼近误差为

(4.10) $ \begin{eqnarray}\label{4.10} \inf\limits_{\vec{v}\in V_h}\|\vec{u}-\vec{u}_h\|_h& \leq &\|\vec{u}-\Pi_h\vec{u}\|_h\\ &=&\sum\limits_{K\in\Gamma_h}(\mu|\vec{u}-\Pi_h\vec{u}|^2_{1, \, K}+(\mu+\lambda)\|\operatorname{div} (\vec{u}-\Pi_h\vec{u})\|^2_{0, \, K})^\frac{1}{2}\\ &\leq& ch(|\vec{u}|_{2, \, \Omega}+\lambda|\operatorname{div} \vec{u}|_{1, \, \Omega}), \end{eqnarray}$

这里的常数$c$与单元剖分尺度$h$无关.

接下来讨论相容误差, $\forall\vec{v}_h\in V_h$有

(4.11) $\begin{eqnarray}\label{4.11}a_h(\vec{u}, \, \vec{v}_h)-\int_{\Omega}\vec{f}\cdot \vec{v}_h{\rm d}x&=&\sum\limits_{K\in\Gamma_h}\int_{K}(\mu\nabla\vec{u}\, :\, \nabla\vec{v}_h+(\mu+\lambda)\operatorname{div}\vec{u}\; \operatorname{div}\vec{v}_h){\rm d}x-\int_{\Omega}\vec{f}\cdot \vec{v}_h{\rm d}x\\&=&\sum\limits_{K\in\Gamma_h}\int_{\partial K}\bigg(\mu \vec{v}_h\frac{\partial\vec{u}}{\partial \vec{n}}+(\mu+\lambda)\operatorname{div}\vec{u}\, \vec{v}_h\, \cdot\, \vec{n}\bigg){\rm d}F, \end{eqnarray}$

设$\vec{w}=\frac{\partial\vec{u}}{\partial \vec{n}}, \, \forall\vec{v}_h\in V_h$,由迹定理,参考单元和一般单元的仿射变换(3.9)及引理4.2,得

(4.12) $\begin{eqnarray}\label{4.12}\sum\limits_{K\in\Gamma_h}\int_{\partial K}\mu \vec{v}_h\frac{\partial\vec{u}}{\partial \vec{n}}{\rm d}F&=&\sum\limits_{K\in\Gamma_h}\sum\limits_{i=1}^5\int_{F_i}\mu(\vec{w}-L(\vec{w}))(\vec{v}_h-P_0(\vec{v}_h)){\rm d}F\\ &\leq &\sum\limits_{K\in\Gamma_h}\sum\limits_{i=1}^5\mu\|\vec{w}-L(\vec{w})\|_{0, \, F_i}\|\vec{v}_h-P_0(\vec{v}_h)\|_{0, \, F_i}\\ &\leq &\sum\limits_{K\in\Gamma_h}\mu h|\vec{w}|_{1, \, K}|\vec{v}_h|_{1, \, K}\\ &\leq &ch|\vec{u}|_{2, \, \Omega}\|\vec{v}_h\|_{h, \, \Omega}. \end{eqnarray} $

(4.13) $\begin{eqnarray}\label{4.13} \sum\limits_{K\in\Gamma_h}\int_{\partial K}\operatorname{div}\vec{u}\, \vec{v}_h\, \cdot\, \vec{n}{\rm d}F&=&\sum\limits_{K\in\Gamma_h}\sum\limits_{i=1}^5\int_{F_i}(\operatorname{div}\vec{u}-L(\operatorname{div}\vec{u}))(\vec{v}_h-P_0\vec{v}_h)\cdot \vec{n}{\rm d}F\\ &\leq &\sum\limits_{K\in\Gamma_h}\sum\limits_{i=1}^5\|\operatorname{div} \vec{u}-L(\operatorname{div}\vec{u})\|_{0, \, F_i}\|\vec{v}_h-P_0\vec{v}_h\|_{0, \, F_i}\\ &\leq& ch|\operatorname{div} \vec{u}|_{1, \, \Omega}\|\vec{v}_h\|_{h, \, \Omega}. \end{eqnarray}$

由(4.11)-(4.13)式得

(4.14) $\begin{equation}\label{4.14} \sup\limits_{\vec{v}\in V_h\setminus \{0\}}\frac{|a_h(\vec{u}, \, \vec{v})-\int_{\Omega}\vec{f}\cdot \vec{v}{\rm d}x|}{\|\vec{v}_h\|_h}\leq ch(|\vec{u}|_{2, \, \Omega}+\lambda|\operatorname{div} \vec{u}|_{1, \, \Omega}). \end{equation} $

综合(4.7), (4.10)和(4.14)式,得误差估计(4.6).

接下来我们证明关于位移的$L^2$模.

定理4.2   在定理4.1的条件下可得位移的$L^2$模估计

(4.15) $\begin{equation}\label{4.15}\|\vec{u}-\vec{u}_h\|_{0, \, \Omega}\leq ch^2(|\vec{u}|_{2, \, \Omega}+\lambda|\operatorname{div} \vec{u}|_{1, \, \Omega}), \end{equation}$

其中$c$是与$h$和$\lambda$无关的常数.

证   方程(2.1)的对偶问题如下

(4.16) $\begin{equation}\label{4.16} \left\{\begin{array}{ll} -\mu\vartriangle\vec{\phi}-(\mu+\lambda) {\rm grad}(\operatorname{div}\vec{\phi}) = \vec{g}, & {\rm in} \ \Omega, \\ \vec{\phi}=0, & {\rm on}\ \partial\Omega, \end{array}\right.\end{equation}$

其中$g\in (L^2(\Omega))^3$且$\phi\in (H^1_0(\Omega))^3\cap (H^2(\Omega))^3$.则相应的变分问题为

(4.17) $ \begin{equation}\label{4.17} a(\vec{v}, \, \vec{\phi})=\int_{\Omega}\vec{g}\cdot\vec{v}{\rm d}x, \quad \forall \vec{v}\in (H^1_0(\Omega))^3, \end{equation}$

这里双线性形$a(\cdot, \, \cdot)$的定义如(2.3)式.由正则性理论可知

(4.18) $ \begin{equation}\label{4.18}\|\vec{\phi}\|_{2, \Omega}+\lambda\|\operatorname{div} \vec{\phi}\|_{1, \Omega}\leq c\|\vec{g}\|_{0, \Omega}.\end{equation}$

对应于(4.17)式的离散问题是求$\vec{\phi}_h\in V_h$满足

(4.19) $ \begin{equation}\label{4.19} a_h(\vec{v}_h, \, \vec{\phi}_h)=\int_{\Omega}\vec{g}\cdot\vec{v}_h{\rm d}x, \quad \forall \vec{v}_h\in V_h.\end{equation}$

由定理4.1的证明可知

(4.20) $\begin{equation}\label{4.20} \|\vec{\phi}-\vec{\phi}_h\|_h\leq ch(|\vec{\phi}|_{2, \Omega}+\lambda|\operatorname{div} \vec{\phi}|_{1, \, \Omega})\leq ch\|\vec{g}\|_{0, \, \Omega}. \end{equation} $

由$L^2$模定义$ \|\vec{u}-\vec{u}_h\|_{0, \Omega}\leq \sup\limits_{\forall \vec{g}\in L^2(\Omega)}\frac{\langle \vec{u}-\vec{u}_h, \, \vec{g}\rangle }{\|\vec{g}\|_{0, \, \Omega}}, $且

(4.21) $\begin{eqnarray}\label{4.22} \langle \vec{u}-\vec{u}_h, \, \vec{g}\rangle &=&a(\vec{u}, \, \vec{\phi})-a_h(\vec{u}_h, \, \vec{\phi}_h)\\ &=&a_h(\vec{u}-\vec{u}_h, \, \vec{\phi}-\Pi_h\vec{\phi})+a_h(\vec{u}-\vec{u}_h, \, \Pi_h\vec{\phi})\\ &&+a_h(\vec{u}_h-\Pi_h\vec{u}_h, \, \vec{\phi}-\vec{\phi}_h)+a_h(\Pi_h\vec{u}_h, \, \vec{\phi}-\vec{\phi}_h). \end{eqnarray} $

下面我们逐项讨论(4.21)式中各部分.由定理4.1和(4.20)式,有

(4.22) $\begin{eqnarray}\label{4.23} a_h(\vec{u}-\vec{u}_h, \, \vec{\phi}-\Pi_h\vec{\phi})&\leq&\|\vec{u}-\vec{u}_h\|_h\|\vec{\phi}-\Pi_h\vec{\phi}\|_h\\ &\leq &ch^2(|\vec{u}|_{2, \, \Omega}+\lambda|\operatorname{div} \vec{u}|_{1, \, \Omega})\|\vec{\phi}\|_{2, \, \Omega}\\ &\leq &ch^2(|\vec{u}|_{2, \, \Omega}+\lambda|\operatorname{div} \vec{u}|_{1, \, \Omega})\|\vec{g}\|_{0, \, \Omega}. \end{eqnarray}$

因为

(4.23) $ \begin{eqnarray}\label{4.24} a_h(\vec{u}-\vec{u}_h, \, \Pi_h\vec{\phi})&=&a_h(\vec{u}, \, \Pi_h \vec{\phi})-a_h(\vec{u}_h, \, \Pi_h \vec{\phi})\\ &=&\sum\limits_{K\in\Gamma_h}\int_K \mu \nabla \vec{u}:\nabla \Pi_h\, \vec{\phi}+(\mu+\lambda)\operatorname{div} \vec{u}\, \operatorname{div} \Pi_h \vec{\phi}{\rm d}x-(f, \Pi_h \vec{\phi})\\ &=&\sum\limits_{K\in\Gamma_h}\bigg[\int_K(-\mu\triangle \vec{u}\, \cdot\, \Pi_h\, \vec{\phi}-(\mu+\lambda){\rm grad}(\operatorname{div} \vec{u})\, \cdot \, \Pi_h \vec{\phi}){\rm d}x\\& &+\int_{\partial K}(\mu\nabla \vec{u}\, \Pi_h \vec{\phi}+(\mu+\lambda)\operatorname{div} \vec{u}\, \Pi_h \vec{\phi})\cdot \vec{n}{\rm d}F\bigg]-(f, \Pi_h \vec{\phi})\\ &=&\sum\limits_{K\in\Gamma_h}\int_{\partial K}\mu\frac{\partial\vec{u}}{\partial\vec{n}}\, \Pi_h\vec{\phi}{\rm d}F+\int_{\partial K}(\mu+\lambda)\operatorname{div} \vec{u}\, \Pi_h \vec{\phi}\cdot\vec{n}{\rm d}F. \end{eqnarray}$

设$\vec{w}=\frac{\partial\vec{u}}{\partial\vec{n}}$,由单元自由度定义可知

(4.24) $ \begin{equation}\label{4.25} \int_{\partial K}L(\vec{w})(\vec{\phi}-\Pi _h \vec{\phi}){\rm d}F=L(\vec{w})\int_{\partial K}(\vec{\phi}-\Pi_h \vec{\phi}){\rm d}F=0. \end{equation} $

根据位移$\vec{u}$跨单元边界的连续性可得

(4.25) $\begin{equation}\label{4.26} \sum\limits_{K\in\Gamma_h}\int_{F_i}\frac{\partial \vec{u}}{\partial \vec{n}}\, \vec{\phi} {\rm d}F=0. \end{equation} $

类似(4.12)和(4.13)式的分析可以推出

(4.26) $ \begin{eqnarray}\label{4.27} \sum\limits_{K\in\Gamma_h}\int_{\partial K}\frac{\partial\vec{u}}{\partial\vec{n}}\, \Pi_h\vec{\phi}{\rm d}F &=&\sum\limits_{K\in\Gamma_h}\int_{\partial K} \bigg(\frac{\partial\vec{u}}{\partial\vec{n}}-L(\frac{\partial\vec{u}}{\partial\vec{n}})\bigg)(\vec{\phi}-\Pi_h\vec{\phi}){\rm d}F\\ &\leq& ch^2|\vec{u}|_{2, \, \Omega}\|\vec{g}\|_{0, \, \Omega}\end{eqnarray} $

和

(4.27) $ \begin{eqnarray}\label{4.28} \sum\limits_{K\in\Gamma_h}\int_{\partial K}\operatorname{div} \vec{u}\, \Pi_h \vec{\phi}\cdot\vec{n}{\rm d}F&=&\sum\limits_{K\in\Gamma_h}\int_{\partial K}(\operatorname{div} \vec{u}-L(\operatorname{div} \vec{u}))\, (\vec{\phi}-\Pi_h \vec{\phi})\cdot\vec{n}{\rm d}F\\ &\leq& ch^2|\operatorname{div} \vec{u}|_{1, \, \Omega}\|\vec{g}\|_{0, \, \Omega}. \end{eqnarray} $

由(4.24)-(4.28)式得

(4.28) $ \begin{equation}\label{4.29} a_h(\vec{u}-\vec{u}_h, \, \Pi_h\vec{\phi})\leq ch^2(|\vec{u}|_{2, \, \Omega}+\lambda|\operatorname{div} \vec{u}|_{1, \, \Omega})\|\vec{g}\|_{0, \, \Omega}\end{equation} $

和

(4.29) $ \begin{equation}\label{4.30} a_h(\vec{\phi}-\vec{\phi}_h, \, \Pi_h\vec{u_h})\leq ch^2(|\vec{u}|_{2, \, \Omega}+\lambda|\operatorname{div} \vec{u}|_{1, \, \Omega})\|\vec{g}\|_{0, \, \Omega}, \end{equation}$

(4.30) $ \begin{eqnarray}\label{4.31} a_h(\vec{u}-\vec{u}_h, \, \vec{\phi}-\vec{\phi}_h)&\leq&\|\vec{u}-\vec{u}_h\|_h\|\phi-\phi_h\|_h\\ &\leq &ch^2(|\vec{u}|_{2, \, \Omega}+\lambda|\operatorname{div} \vec{u}|_{1, \, \Omega})(|\vec{\phi}|_{2, \, \Omega}+\lambda|\operatorname{div} \vec{\phi}|_{1, \, \Omega})\\&\leq &ch^2(|\vec{u}|_{2, \, \Omega}+\lambda|\operatorname{div} \vec{u}|_{1, \, \Omega})\|\vec{g}\|_{0, \, \Omega}. \end{eqnarray} $

通过上述逐项分析得到(4.21)-(4.22), (4.28)-(4.30)式,代入$L^2$模定义可得结论.