二维有限元的类型主要是三角形元和矩形元,相应地,三维有限元的类型主要是四面体元和立方体元.这些种类的单元在各种问题中的应用已有很多研究,可以参考文献[1-5, 15].三维有限元中有一类特殊单元即三棱柱单元,它对截面复杂的柱体区域比四面体元和立方体元有更好的适应性,因而有很好的应用前景.但研究的并不多,三棱柱单元的上下底面是三角形,三个侧面是矩形,构造有一定难度.三棱柱单元最近的研究可以参考文献[6-8].

二阶椭圆问题混合有限元的研究已经有很多^[9],例如Raviart-Thomas,三角形元和矩形元^[10],四面体元和立方体元^[11], BDM^[12]元和BDDF^[13]元.包括三角形元、四面体元、矩形元、立方体元.但对三棱柱元的研究并不多,对截面复杂的柱体区域,三棱柱元比四面体元和立方体元都具有优势,文献[14]中构造了散度空间的三棱柱协调元,由于其构造过于复杂,没有在二阶混合问题中得到应用,该文构造一个形式简单的低阶三棱柱元,证明了它的正定性,收敛性,给出了误差估计,达到最优阶.

设$ \Omega \subset \mathbb{R}^{3} $是有界的凸多边形区域, $ \partial\Omega $为其边界, $ f \in L^{2}(\Omega) $为已知函数.

考虑二阶椭圆混合变分问题^[9]:求$ (\underline{\psi}, u)\in H \times M $,使得

(2.1) $ \begin{equation} \left \{ \begin{array}{ll} a(\underline{\psi}, \underline{\varphi}) + b(\underline{\varphi}, u ) = 0, & \forall \underline{\varphi}\ \in H, \\ b(\underline{\psi}, \nu) = G(\nu), & \forall \nu \in M, \end{array} \right. \end{equation} $

其中

$ a(\underline{\psi}, \underline{\varphi}) = \int_{\Omega}\underline{\psi}\cdot\underline{\varphi}, b(\underline{\varphi}, u) = \int_{\Omega}u \text{div}\underline{\varphi}, G(\nu) = -\int_{\Omega}f\nu, $

$ H = H(\text{div}) = \{{\underline{\varphi}\in L^{2}(\Omega)^{3}; \text{div}\underline{\varphi}\in L^{2}(\Omega)}\}, M = L^{2}(\Omega), $

$ \| \underline{\varphi} \|_{H} = (\| \underline{\varphi} \|^{2}_{0}+\| \text{div} \underline{\varphi} \|^{2}_{0})^{\frac{1}{2}}, $

$ a(\cdot, \cdot) $, $ b(\cdot, \cdot) $分别是$ H\times H $, $ H\times M $上连续双线性型. $ G(\cdot) $是连续线性型, $ f \in L^{2}(\Omega). $

引理1^[9]  因为混合变分问题(2.1)满足如下条件:

(1) $ a(\cdot, \cdot) $在$ Z \times Z $上是正定的,即存在常数$ \alpha > 0 $,使得

$ a(v, v)\geq \alpha \|v \|^{2}_{H}, \forall v \in Z, $

其中$ Z = \{ v\in H; b(v, q) = 0, \forall q \in M \} $是$ H $的闭子空间;

(2) $ b(\cdot, \cdot) $满足$ BB $条件,即存在$ \beta > 0 $,使得

$ \sup\limits_{v \in H }\frac{b( v, q )}{\| v \|_{H}}\geq \beta \| q \|^{2}_{ M }, \forall q \in M. $

则混合变分问题(2.1)存在唯一解.

将区域$ \Omega $剖分成有限个子区域$ T $,剖分记为$ T_{h} $,设$ H_{h} $, $ M_{h} $分别是$ H $, $ M $的逼近空间,如果$ H_{h}\subset H, M_{h} \subset M $,则称为协调元空间,否则称为非协调元空间.

对于协调元空间,则混合问题(2.1)的离散问题为:求$ (\underline{\psi_{h}}, u_{h})\in H_{h} \times M_{h} $,使得

(2.2) $ \begin{equation} \left \{ \begin{array}{ll} a(\underline{\psi_{h}}, \underline{\varphi_{h}}) + b(\underline{\varphi_{h}}, u _{h}) = 0, & \forall \underline{\varphi_{h}}\ \in H_{h} , \\ b(\underline{\psi_{h}}, \nu_{h}) = G(\nu_{h}), & \forall \nu_{h} \in M_{h}. \end{array} \right . \end{equation} $

引理2^[9]  如果离散问题(2.2)满足如下条件:

(1) $ a(\cdot, \cdot) $在$ Z_{h}\times Z_{h} $上强制,即存在常数$ \alpha > 0 $,使得

$ a(v_{h}, v_{h})\geq \alpha \|v_{h} \|^{2}_{H_{h}}, \forall v_{h} \in Z_{h}, $

其中$ Z_{h} = \{ v_{h}\in H_{h}; b(v_{h}, q_{h}) = 0, \forall q_{h} \in M_{h} \} $是$ H_{h} $的闭子空间;

(2) $ b(\cdot, \cdot) $连续且满足离散的$ BB $条件,即存在$ \beta > 0 $,使得

$ \sup\limits_{v_{h} \in H_{h} }\frac{b( v_{h}, q_{h} )}{\| v_{h} \|_{H_{h}}}\geq \beta \| q_{h} \|^{2}_{ M_{h} }, \forall q_{h} \in M_{h}. $

则离散问题(2.2)存在唯一解,并且下面的误差估计成立

(2.3) $ \begin{equation} \begin{array}{l} \| \underline{\psi} - \underline{\psi_{h}} \|_{H} + \| u - u_{h} \|_{M} \leq c \Big (\inf\limits_{\underline{\varphi_{h}} \in H_{h} }\| \underline{\psi} - \underline{\varphi_{h}} \|_{H} + \inf\limits_{ \nu_{h} \in M_{h} }\| u - \nu_{h}\|_{M}\Big). \end{array} \end{equation} $

设$ T_{h} $是$ \Omega $的三棱柱剖分, $ \Omega = \bigcup\limits_{T\in T_{h}}T $是一个柱体区域, $ T $是其中一个三棱柱,满足网格的正则性条件.

设参考元$ \hat{T} $如下图, 6个顶点分别为

记三棱柱的5个面分别为$ \widehat{S}_{i} $, $ 1\leq i \leq 5 $,其中四边形$ \hat{a}_{1}\hat{a}_{2}\hat{a}_{5}\hat{a}_{4} $记为$ \widehat{S}_{1} $,四边形$ \hat{a}_{1}\hat{a}_{3}\hat{a}_{6}\hat{a}_{4} $记为$ \widehat{S}_{2} $,三边形$ \hat{a}_{4}\hat{a}_{5}\hat{a}_{6} $记为$ \widehat{S}_{3} $,三边形$ \hat{a}_{1}\hat{a}_{2}\hat{a}_{3} $记为$ \widehat{S}_{4} $,四边形$ \hat{a}_{2}\hat{a}_{3}\hat{a}_{6}\hat{a}_{5} $记为$ \widehat{S}_{5} $.

$ \widehat{T} $的形函数空间定义为

$ P(\widehat{T}) = \widehat{P}^{3}_{0}+ \left( \begin{array}{rrr} \hat{x} \\ \hat{y} \\ \hat{z} \\ \end{array} \right) \widehat{P}_{0} + \left( \begin{array}{rrr} \hat{x}\\ \hat{y}\\ -2\hat{z} \\ \end{array} \right) \widehat{P}_{0}, $

其中$ \dim P(\widehat{T}) = 5 $, $ P(\widehat{T}) $中元素$ \widehat{\underline{\varphi}} = (\hat{\varphi}_{1}, \hat{\varphi}_{2}, \hat{\varphi}_{3}) $的自由度$ \widehat{\Sigma} $定义为

$ \int_{\hat{S}} \underline{\hat{\varphi}} \cdot \underline{\hat{n}}, \forall \widehat{S} \subset \partial \hat{T}. $

下面证明单元构造的适定性.

引理3  自由度能唯一确定形函数空间$ P (\widehat{T}) $中的元素,即单元构造是适定的.

证  只需证明,当$ \widehat{\underline{\varphi}}\in P(\widehat{T}) $,且$ \widehat{\underline{\varphi}} $的自由度为0,则$ \widehat{\underline{\varphi}} = 0 $,设

$ \int_{\widehat{S}} \underline{\hat{\varphi}} \cdot \underline{\hat{n}} = 0, \widehat{S} \subset \partial \widehat{T}. $

由于$ \widehat{\underline{\varphi}}\in P(\widehat{T}) $,设$ \widehat{\underline{\varphi}} = \left(\begin{array}{c} a+\alpha \hat{x}+ \beta \hat{x} \\ b+ \alpha \hat{y}+ \beta \hat{y} \\ c+ \alpha \hat{z}-2 \beta \hat{z} \end{array} \right) $,其中$ a, b, c, \alpha, \beta $均为常数.

设$ \underline{\hat{n}}^{(i)} $是面$ \widehat{S}_{i} $上单位外法向量, $ 1\leq i \leq 5 $,则有

$ \begin{eqnarray*} \left \{ \begin{array}{ll} \widehat{S_{1}}上\hat{y} = 0, &\underline{\hat{n}}^{(1)} = (0, -1, 0)^{T}, \\ \widehat{S_{2}}上\hat{x} = 0, & \underline{\hat{n}}^{(2)} = (-1, 0, 0)^{T}, \\ \widehat{S_{3}}上\hat{z} = 1, & \underline{\hat{n}}^{(3)} = (0, 0, 1)^{T}, \\ \widehat{S_{4}}上\hat{z} = 0, &\underline{\hat{n}}^{(4)} = (0, 0, -1)^{T}, \\ \widehat{S_{5}}上\hat{x}+\hat{y} = 1, & \underline{\hat{n}}^{(5)} = \frac{1}{\sqrt{2}}(1, 1, 0)^{T}. \end{array} \right. \end{eqnarray*} $

通过计算,可以得到

$ \begin{eqnarray*} \underline{\widehat{\varphi}} \cdot \underline{\hat{n}}^{(1)}|_{\widehat{S_{1}}} & = & -(b+ \alpha \hat{y}+ \beta \hat{y}) = -b , \\ \underline{\widehat{\varphi}} \cdot \underline{\hat{n}}^{(2)}|_{\widehat{S_{2}}} & = & -(a+\alpha \hat{x}+ \beta \hat{x}) = -a , \\ \underline{\widehat{\varphi}} \cdot \underline{\hat{n}}^{(3)}|_{\widehat{S_{3}}} & = & c+ \alpha \hat{z}-2 \beta \hat{z} = c+\alpha- 2\beta , \\ \underline{\widehat{\varphi}} \cdot \underline{\hat{n}}^{(4)}|_{\widehat{S_{4}}} & = & -( c+ \alpha \hat{z}-2 \beta \hat{z }) = - c , \\ \underline{\widehat{\varphi}} \cdot \underline{\hat{n}}^{(5)}|_{\widehat{S_{5}}} & = & a+\alpha \hat{x}+ \beta \hat{x} + b+ \alpha \hat{y}+ \beta \hat{y} = a+\alpha + \beta + b, \end{eqnarray*} $

即$ \underline{\widehat{\varphi}} \cdot \underline{\hat{n}}^{(i)}|_{\widehat{S_{i}}} = $常数, $ 1\leq i \leq 5. $

意味着

(3.1) $ \begin{eqnarray} \underline{\widehat{\varphi}} \cdot \underline{\hat{n}}^{(i)}|_{\widehat{S_{i}}} = 0, 1\leq i \leq 5, \end{eqnarray} $

即可得

$ a = b = c = \alpha = \beta = 0, $

故$ \widehat{\underline{\varphi}} = 0 $,引理证明完毕.

从参考元$ \widehat{T} $到一般单元$ T $,如下图:

一般单元$ T $的6个顶点分别为$ a_{1}(x_{1}, y_{1}, z_{1}), a_{2}(x_{2}, y_{2}, z_{1}), a_{3}(x_{3}, y_{3}, z_{1}), $$ a_{4}(x_{1}, y_{1}, z_{1}+h_{3}), $$ a_{5}(x_{2}, y_{2}, z_{1}+ h_{3}), a_{6}(x_{3}, y_{3}, z_{1} + h_{3}) $,三棱柱的5个面分别记为$ S_{i}, 1\leq i \leq 5 $.

四边形$ {a}_{1}{a}_{2}{a}_{5}{a}_{4} $记为$ {S}_{1} $,四边形$ {a}_{1}{a}_{3}{a}_{6}{a}_{4} $记为$ {S}_{2} $,三边形$ {a}_{4}{a}_{5}{a}_{6} $记为$ {S}_{3} $,三边形$ {a}_{1}{a}_{2}{a}_{3} $记为$ {S}_{4} $,四边形$ {a}_{2}{a}_{3}{a}_{6}{a}_{5} $记为$ {S}_{5} $.

从参考元到一般单元的变换记为

$ F:\widehat{T}\rightarrow T, $

$ \underline{ X } = F( \underline{\widehat{X}}) = B \underline{\widehat{X}}+ b. $

其中

$ 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 & h \\ \end{array} \right) ,~ b = \left( \begin{array}{rrr} x_{1} \\ y_{1} \\ z_{1}\\ \end{array} \right). $

从参考元到一般单元,标量函数记为

$ \nu(x) = \nu( F (\underline{ \hat{x}})) = \hat{\nu}( \hat{ x } ). $

对于向量函数,采用Piola变换,由$ \underline{x} = F (\underline{ \hat{x}}) $,有

$ \underline{\varphi}(\underline{x}) = (\det B)^{-1} B \underline{\widehat{\varphi}}(\underline{\hat{x}}). $

相对于参考元,一般单元$ T $上的形函数空间定义为

(3.2) $ \begin{eqnarray} P({T}) = P^{3}_{0}+ \left( \begin{array}{rrr} {x} \\ {y} \\ {z} \\ \end{array} \right) P_{0} + \left( \begin{array}{rrr} {x} \\ {y} \\ -2{z} \\ \end{array} \right) P_{0}, \end{eqnarray} $

其中$ P({T}) $上元素的自由度$ \Sigma_{T} $定义为

(3.3) $ \begin{eqnarray} \int_{{S}} \underline{{\varphi}} \cdot \underline{{n}}, \forall{S} \subset \partial {T}. \end{eqnarray} $

引理4  有限元$ (T, P(T), \Sigma_{T}) $和$ (\widehat{T}, P(\widehat{T}), \widehat{\Sigma}) $在Piola变换下仿射等价,所以自由度(3.3)能唯一确定$ P({T}) $中的元素.

证  在仿射变换下,显然$ \widehat{T}\longleftrightarrow T, $

$ \forall \underline{\widehat{\varphi}} \in P(\widehat{T}), \underline{\widehat{\varphi}} = \widehat{P}^{3}_{0}+ \left( \begin{array}{rrr} \hat{x} \\ \hat{y} \\ \hat{z} \\ \end{array} \right) \widehat{P}_{0} + \left( \begin{array}{rrr} \hat{x} \\ \hat{y}\\ -2\hat{z} \\ \end{array} \right) \widehat{P}_{0}. $

设$ B \underline{\hat{x}} = \underline{x}-\underline{b} $,可以得到

$ \begin{eqnarray*} \underline{\varphi}(x)& = & (\det B)^{-1} B \underline{\widehat{\varphi}} = (\det B)^{-1}( B \widehat{P}^{3}_{0}+ B \left( \begin{array}{rrr} \hat{x} \\ \hat{y} \\ \hat{z} \\ \end{array} \right) \widehat{P}_{0} + B \left( \begin{array}{cc} \hat{x} \\ \hat{y} \\ -2\hat{z} \\ \end{array} \right) \widehat{P}_{0})\\ & = & (\det B)^{-1}( B \widehat{P}^{3}_{0}+ \left( \begin{array}{c} {x} \\ {y} \\ {z} \\ \end{array} \right) \widehat{P}_{0}-\underline{b}\widehat{P}_{0} + B \left( \begin{array}{c} \hat{x} \\ \hat{y} \\ -2\hat{z} \\ \end{array} \right) \widehat{P}_{0})\\ & = & (\det B)^{-1}( B \widehat{P}^{3}_{0}+ \left( \begin{array}{c} {x} \\ {y} \\ {z} \\ \end{array} \right) \widehat{P}_{0}-\underline{b}\widehat{P}_{0} + \left( \begin{array}{c} {x} \\ {y} \\ -2{z} \\ \end{array} \right) \widehat{P}_{0}+ \left( \begin{array}{rrr} I_{2} & 0 \\ 0 & -2 \\ \end{array} \right)(-\underline{b})\widehat{P}_{0}) \\ & = & (\det B)^{-1}( B \widehat{P}^{3}_{0}-\underline{b}\widehat{P}_{0}+ \left( \begin{array}{ cc} I_{2} & 0 \\ 0 & -2 \\ \end{array} \right)(-\underline{b})\widehat{P}_{0}+ \left( \begin{array}{c} {x} \\ {y} \\ {z} \\ \end{array} \right) \widehat{P}_{0} + \left( \begin{array}{c} {x}\\ {y} \\ -2{z} \\ \end{array} \right) \widehat{P}_{0} )\\ &\in & P(T). \end{eqnarray*} $

反之也正确.

因此$ P(\widehat{T}) $仿射等价于$ P({T}) $,即

$ P(\widehat{T}) \longleftrightarrow P({T}). $

下面考虑自由度.

假设$ \underline{{n}}^{(i)} $是面$ {S}_{i} $上的外法向量, $ \mid S_{i} \mid $是面$ {S}_{i} $的面积, $ 1\leq i \leq 5 $,

$ \begin{eqnarray*} \left \{ \begin{array}{ll} S_{1}上, x(y_{2}-y_{1})-y(x_{2}-x_{1}) = \delta_{1}, \\ S_{2}上, x(y_{3}-y_{1})-y(x_{3}-x_{1}) = \delta_{2}, \\ S_{3}上, z = 1, \\ S_{4}上, z = 0, \\ S_{5}上, x(y_{3}-y_{2})-y(x_{3}-x_{2}) = \delta_{5}, \end{array} \right . \end{eqnarray*} $

其中

$ \begin{eqnarray*} &&\delta_{1} = x_{1}y_{2} - x_{2}y_{1}, \delta_{2} = x_{1}y_{3} - x_{3}y_{1}, \delta_{5} = x_{2}y_{3} - x_{3}y_{2}. \\ &&\underline{{n}}^{(1)} = \frac{1}{\sqrt{(y_{2}-y_{1})^{2} + (x_{2}-x_{1})^{2}}}(y_{2}-y_{1}, -(x_{2}-x_{1}), 0)^{T}, \\ &&\underline{{n}}^{(2)} = \frac{1}{\sqrt{(y_{3}-y_{1})^{2} + (x_{3}-x_{1})^{2}}}(-(y_{3}-y_{1}), x_{3}-x_{1}, 0)^{T}, \\ && \underline{{n}}^{(3)} = (0, 0, 1)^{T}, \\ && \underline{{n}}^{(4)} = (0, 0, -1)^{T}, \\ &&\underline{{n}}^{(5)} = \frac{1}{\sqrt{(y_{3}-y_{2})^{2} + (x_{3}-x_{2})^{2}}}(y_{3}-y_{2}, x_{3}-x_{2}, 0)^{T}.\\ \end{eqnarray*} $

显然

(3.4) $ \begin{equation} \begin{array}{l} \underline{\varphi}(x)\cdot \underline{n}^{i}|_{S_{i}} =常数, 1\leq i \leq 5. \end{array} \end{equation} $

由于

$ B^{-T} = \frac{1}{\det B}\left(\begin{array}{rrr} h(y_{3}-y_{1}) & \ -h(y_{2}-y_{1})\ & 0 \\ -h(x_{3}-x_{1}) & h(x_{2}-x_{1}) & 0 \\ 0 & 0 & (x_{2}-x_{1})(y_{3}-y_{1})- (y_{2}-y_{1})(x_{3}-x_{1})\\ \end{array} \right), $

则有

$B^{-T}\underline{\widehat{n}}^{(1)} = B^{-T}\left(\begin{array}{c} 0 \\ -1 \\ 0 \end{array} \right) = \frac{1}{\det B} \left(\begin{array}{c} h(y_{2}-y_{1})\\ -h(x_{2}-x_{1})\\ 0 \end{array}\right) = \frac{1}{\det B}\frac{\mid S_{1} \mid }{\mid \widehat{S}_{1} \mid}\underline{{n}}^{(1)},$

即

$\underline{{n}}^{(1)} = \bigg(\frac{\mid \widehat{S}_{1}\mid}{\mid S_{1} \mid} \det B \bigg)B^{-T}\underline{\widehat{n}}^{(1)}.$

同理可得,对$ i = 2, 3, 4, 5 $上式也成立,即得到

$ \underline{{n}}^{(i)} = \bigg(\frac{\mid \widehat{S}_{i}\mid}{\mid S_{i} \mid}\det B \bigg)B^{-T}\underline{\widehat{n}}^{(i)}, 1 \leq i \leq 5, $

(3.5) $ \begin{eqnarray} \int_{S_{i}} \underline{\varphi}\underline{{n}}^{(i)} & = & \frac{\mid {S}_{i}\mid}{\mid \widehat{S}_{i} \mid}\int_{\widehat{S}_{i}}(\det B )^{-1} B \underline{\widehat{\varphi}} \cdot \bigg (\frac{\mid \widehat{S}_{i}\mid}{\mid S_{i} \mid} \det B \bigg) B^{-T}\underline{\widehat{n}}^{(i)} \\ & = & \int_{\widehat{S}_{i}}(B^{-T} B )\underline{\widehat{\varphi}} \cdot \underline{\widehat{n}}^{(i)} \\ & = & \int_{\widehat{S}_{i}}\underline{\widehat{\varphi}} \cdot \underline{\widehat{n}}^{(i)}, 1 \leq i \leq 5 . \end{eqnarray} $

进而可以推断出$ P(T) $上的自由度和$ P(\widehat{T}) $上自由度仿射等价.

因此,任意有限元$ (T, P(T), \Sigma_{T}) $仿射等价于$ (\widehat{T}, P(\widehat{T}), \widehat{\Sigma}) $,引理证明完毕.

由单元构造,相应的有限元空间$ H_{h} $, $ M_{h} $定义为

$ \begin{eqnarray*} \left \{ \begin{array}{ll} H_{h} = \{\underline{\varphi}_{h}\mid_{T}\in P(T), [\underline{\varphi}_{h}\cdot \underline{n} ]\mid_{S} = 0, \forall S \subset \partial T , \forall T \ \in T_{h}\} , \\ M_{h} = \{\nu_{h}\in L^{2}(\Omega); \nu_{h}\in P_{0}(T), \ \ \ \ \forall T \ \in T_{h}\}. \end{array} \right . \end{eqnarray*} $

在空间$ H_{h} $定义中,只考虑内测面$ S $,结合(3.1), (3.3), (3.4)和(3.5)式,可知$ \underline{\varphi}_{h}\cdot \underline{n} $跨过单元边界连续,即$ [\underline{\varphi}_{h}\cdot \underline{n}]\mid_{S} = 0, \forall S \subset \partial T, \forall T \ \in T_{h} $,意味着

$ H_{h}\subset H(\text{div}), $

又显然$ M_{h}\subset M = L^{2}(\Omega) $,则$ H_{h} $和$ M_{h} $对于二阶问题(2.1)是协调元空间,在离散格式(2.2)中,由于$ \forall \underline{\varphi}_{h} \in H_{h}, \text{div}\underline{\varphi}_{h}\mid_{T} \in P_{0}, $所以

$ \text{div}\underline{\varphi}_{h}\in M_{h}, $

那么

$ \begin{eqnarray*} Z_{h} & = & \{ \underline{\varphi}_{h} \in H_{h}; b( \underline{\varphi}_{h}, \nu_{h} ) = 0, \forall \nu_{h}\in M_{h} \}\\ & = &\bigg \{ \underline{\varphi}_{h} \in H_{h}; \int_{\Omega}\nu_{h}\cdot \text{div} \underline{\varphi}_{h} = 0, \forall \nu_{h}\in M_{h} \bigg\}\\ & = & \{ \underline{\varphi}_{h} \in H_{h}; \text{div} \underline{\varphi}_{h} = 0 \} \subset Z, \end{eqnarray*} $

因此得出$ a(\cdot, \cdot) $在$ Z_{h} $上强制.

定义插值算子$ \widehat{\Pi}:H^{1}(\hat{T})\rightarrow P(\hat{T}) $和$ \Pi_{T}: H^{1}(T)\rightarrow P(T) $分别满足

$ \int_{\widehat{S}} ( \underline{\hat{\varphi}} - \widehat{\Pi} \underline{\hat{\varphi}}) \cdot \underline{\hat{n}} = 0, \forall \widehat{S}\subset\partial \widehat{T}, \int_{S} ( \underline{\varphi} - \Pi_{T} \underline{\varphi}) \cdot \underline{n} = 0, \forall S \subset \partial T. $

由引理4可知,在$ Piola $变换下$ \Pi_{T} $与$ \widehat{\Pi} $仿射等价,即

$ \Pi_{T}\underline{\varphi} = (\det B )^{-1} B \widehat{\Pi}\underline{\hat{\varphi}}. $

由文献[1]中定理2.5知

$ \| \underline{\hat{v}\hat{n}}\|_{-\frac{1}{2}, \partial \widehat{T}} \leq \| \underline{\hat{v}}\|_{H(\text{div})}, $

因此

(4.1) $ \begin{eqnarray} \|\widehat{\Pi} \underline{\hat{\nu }}\|_{H(\text{div})} \leq \hat{c}\sum \limits_{ \hat{l}_{i} \subset \partial \hat{T} } \bigg| \int_{\hat{l}_{i}} \underline{\hat{\nu }} \cdot \underline{\hat{n}}\bigg| \leq \hat{c} \| \underline{\hat{v}\hat{n}} \|_{-\frac{1}{2}, \partial \hat{T}} \leq \hat{c} \| \underline{\hat{v}}\|_{H(\text{div})}. \end{eqnarray} $

同时

(4.2) $ \begin{eqnarray} \text{div}\underline{\widehat{\varphi}} = \widehat{\bigtriangledown}\underline{\widehat{\varphi}} = B^{T}\bigtriangledown\cdot\underline{\widehat{\varphi}} = \bigtriangledown \cdot B \underline{\widehat{\varphi}} = (\det B)\bigtriangledown\cdot\underline{\varphi} = (\det B)\text{div}\underline{\varphi}, \end{eqnarray} $

整体插值算子$ \Pi_{h}:H(\text{div})\rightarrow H_{h} $定义为: $ \Pi_{h}\mid_{T} = \Pi_{T}, \forall T \in T_{h}. $

根据(4.1)和(4.2)式,借助Piola尺度变换,即可推断出

(4.3) $ \begin{eqnarray} \|\Pi_{h}\underline{\varphi}\|_{H(\text{div})}\leq c \|\underline{\varphi}\|_{H(\text{div})}. \end{eqnarray} $

引理5   $ b(\cdot, \cdot) $满足离散$ BB $条件.

证  利用插值条件和Green公式,可以得到

(4.4) $ \begin{eqnarray} b(\underline{\varphi}- \Pi_{T} \underline{\varphi}, q_{h}) & = & \sum \limits_{ T }\int_{T} q_{h} \cdot \text{div}(\underline{\varphi}- \Pi_{T}\underline{\varphi})\\ & = & \sum \limits_{ T }\int_{\partial T} q_{h} (\underline{\varphi}- \Pi_{T}\underline{\varphi})\cdot \underline{n}-\sum \limits_{ T }\int_{T} (\underline{\varphi}- \Pi_{T}\underline{\varphi})\cdot \bigtriangledown q_{h}\\ & = & \sum \limits_{ T }q_{h}\int_{\partial T} (\underline{\varphi}- \Pi_{T}\underline{\varphi})\cdot \underline{n}-0 = 0, \forall q_{h} \in M_{h}. \end{eqnarray} $

由(4.3)式, (4.4)式和Fortin准则^[9],引理容易证明.

定理1  假设$ (\underline{\psi}, u) $和$ (\underline{\psi}_{h}, u_{h}) $分别是问题(2.1)和(2.2)的解,则存在一个常数c,使得下面的误差估计成立:

$ \|\underline{\psi} - \underline{\psi}_{h} \|_{H(\text{div})}+ \| u - u_{h} \|_{0} \leq ch (|\underline{\psi}|_{1} + |\text{div} \underline{\psi}|_{1} + | u |_{1}), $

其中$ h = \max h_{T} $, $ h_{T} $是$ T $的直径, $ c $是一个与$ h $无关的常数.

证  已知$ a(\cdot, \cdot) $在$ Z_{h} $上强制,由引理5知$ b(\cdot, \cdot) $满足离散$ BB $条件,由混合元理论^[9]和方程(2.3)得

$ \| \underline{\psi} - \underline{\psi_{h}} \|_{H(\text{div})} + \| u - u_{h} \|_{0} \leq c \bigg (\inf\limits_{\underline{\varphi_{h}} \in H_{h} }\| \underline{\psi} - \underline{\varphi_{h}} \|_{H(\text{div})} + \inf\limits_{ \nu_{h} \in M_{h} }\| u - \nu_{h}\|_{0}\bigg). $

取

$ \underline{\varphi_{h}} = \Pi_{h}\underline{\psi}, \nu_{h} = I_{h}u, $

其中$ \Pi_{h} $和$ I_{h} $分别是$ H_{h} $和$ M_{h} $的插值算子,则有

$ \| \underline{\psi} - \underline{\psi_{h}} \|_{H(\text{div})} + \| u - u_{h} \|_{0} \leq c (\| \underline{\psi} - \Pi_{h}\underline{\psi} \|_{H(\text{div})} + \| u - I_{h}u \|_{0}). $

因为$ H_{h} $和$ M_{h} $包含完整的零次多项式,借助于插值理论,可以得到

$ \| \underline{\psi} - \Pi_{h}\underline{\psi} \|_{0}\leq ch|\underline{\psi}|_{1},~\| u - I_{h}u \|_{0}\leq ch|u|_{1}. $

由于$ \text{div} \Pi_{h}\underline{\psi}\in P_{0} $,由插值条件

$ \int_{\Omega}\text{div} \Pi_{h}\underline{\psi} = \sum\limits_{T}\int_{\partial T }\Pi_{h}\underline{\psi}\cdot \underline{n} = \sum\limits_{T}\int_{\partial T }\underline{\psi}\cdot \underline{n} = \int_{\Omega}\text{div} \underline{\psi}, $

所以

$ \text{div} \Pi_{h}\underline{\psi} = P_{h} \text{div} \underline{\psi}, $

其中$ P_{h}:L^{2}(\Omega)\rightarrow M_{h} $是$ L^{2} $上的投影算子,由此可得

$ \| \text{div}\underline{\psi} - \text{div}\Pi_{h}\underline{\psi} \|_{0} = \| \text{div}\underline{\psi} - P_{h}\text{div} \underline{\psi} \|_{0} \leq ch|\text{div}\underline{\psi}|_{1}, $

因此

$ \| \underline{\psi} - \Pi_{h}\underline{\psi} \|_{H(\text{div})} \leq ch(|\underline{\psi}|_{1} + |\text{div}\underline{\psi}|_{1}), $

定理证明完毕.