The Lower/Upper Bound Property of Approximate Eigenvalues by Nonconforming Finite Element Methods for Elliptic Operators

Abstract

This paper is a complement of the work (Hu et al. in arXiv:1112.1145v1[math.NA], 2011), where a general theory is proposed to analyze the lower bound property of discrete eigenvalues of elliptic operators by nonconforming finite element methods. One main purpose of this paper is to propose a novel approach to analyze the lower bound property of discrete eigenvalues produced by the Crouzeix–Raviart element when exact eigenfunctions are smooth. In particular, under some conditions on the triangular mesh, it is proved that the Crouzeix–Raviart element method of the Laplace operator yields eigenvalues below exact ones. Such a theoretical result explains most of numerical results in the literature and also partially answers the problem of Boffi (Acta Numerica 1–120, 2010). This approach can be applied to the Crouzeix–Raviart element of the Stokes eigenvalue problem and the Morley element of the buckling eigenvalue problem of a plate. As a second main purpose, a new identity of the error of eigenvalues is introduced to study the upper bound property of eigenvalues by nonconforming finite element methods, which is successfully used to explain why eigenvalues produced by the rotated $Q_1$ element of second order elliptic operators (when eigenfunctions are smooth), the Adini element (when eigenfunctions are singular) and the new Zienkiewicz-type element of fourth order elliptic operators, are above exact ones.

Appendix

Let $T = T_{a,\theta ,h}$ and ${\hat{T}}=T_{1,\frac{\pi }{2},1}$. We introduce the affine transformation $F:{\hat{T}}\rightarrow T_{a,\theta ,h}$ by

$$\begin{aligned} x:=F({\hat{x}}):=B_{a,\theta ,h}{\hat{x}}, \end{aligned}$$

where

On element $T$, the quadratic polynomial $\Pi _2 u-u_h$ can be expressed as

$$\begin{aligned} (\Pi _2 u-u_h)|_T=\varvec{c}^\top {\varvec{\phi }}, \end{aligned}$$

where $\varvec{c}=(c_1\quad c_2\quad c_3\quad c_4\quad c_5\quad c_6)^\top$ and ${\varvec{\phi }}=(1\quad x\quad y\quad x^2\quad xy\quad y^2)^\top$ for some parameters $c_i\,, i=1,\ldots , 6$. The area of $T$ is $S=\frac{1}{2}ah^2\sin \theta$. On one hand,

$$\begin{aligned} ||\Delta (\Pi _2 u-u_h)||_{0,T}^2=\int \limits _T(\Delta \varvec{c}^\top {\varvec{\phi }})^2dxdy =\varvec{c}^\top \bigg (\int \limits _T\Delta {\varvec{\phi }}\Delta {\varvec{\phi }}^\top dxdy\bigg )\varvec{c}=\varvec{c}^\top M_2\varvec{c}. \end{aligned}$$

where $M_2$ is a $6\times 6$ matrix which has only $4$ nonzero elements, i.e., $M_2(4,4)=M_2(6,6)=4S,M_2(4,6)=M_2(6,4)=4S$. On the other hand,

$$\begin{aligned} |T|^{-1}\bigg (\int \limits _T(\Pi _1-I)(\Pi _2u-u_h)dxdy\bigg )^2&= |T|^{-1}\bigg (\int \limits _T(\Pi _1-I)\varvec{c}^\top {\varvec{\phi }}dxdy\bigg )^2\\&= |T|^{-1}\bigg (\int \limits _T(\Pi _1\!-\!I)(c_4{\varvec{\phi }}(4)\!+\!c_5{\varvec{\phi }}(5) +c_6{\varvec{\phi }}(6))dxdy\bigg )^2\\&= \varvec{c}^\top M_3\varvec{c}, \end{aligned}$$

where $M_3$ is also a $6\times 6$ matrix which has $9$ nonzero components as follows

$$\begin{aligned} M_3(4,4)&= (ah^6\sin \theta (a^2\cos ^2\theta - a\cos \theta + 1)^2)/648,\\ M_3(4,5)&= M_3(5,4)= -(a^2h^6\sin \theta (\sin \theta - a\sin (2\theta ))(a^2\cos ^2\theta - a\cos \theta + 1))/1296,\\ M_3(4,6)&= M_3(6,4)=(a^3h^6\sin ^3\theta (a^2\cos ^2\theta - a\cos \theta + 1))/648,\\ M_3(5,5)&= (a^3h^6\sin \theta (\sin \theta - a\sin (2\theta ))^2)/2592,\\ M_3(5,6)&= M_3(6,5)= -(a^4h^6\sin ^3\theta (\sin \theta - a\sin (2\theta )))/1296,\\ M_3(6,6)&= (a^5h^6\sin ^5\theta )/648. \end{aligned}$$

Finally,

$$\begin{aligned} |\Pi _2 u-u_h|^2_{1,T}&= \int \limits _T\left( \left( \frac{\partial \varvec{c}^\top {\varvec{\phi }}}{\partial x}\right) ^2 +\left( \frac{\partial \varvec{c}^\top {\varvec{\phi }}}{\partial y}\right) ^2\right) dxdy\\&= \int \limits _{\hat{T}}\frac{1}{h^2\sin ^2\theta } \left( \left( \frac{\partial \widehat{{\varvec{c}}^\top {\varvec{\phi }}}}{\partial {\hat{x}}}\right) ^2 -\frac{2\cos \theta }{a}\frac{\partial \widehat{{\varvec{c}}^\top {\varvec{\phi }}}}{\partial {\hat{x}}} \frac{\partial \widehat{{\varvec{c}}^\top {\varvec{\phi }}}}{\partial {\hat{y}}}\!+\!\frac{1}{a^2} \left( \frac{\partial \widehat{{\varvec{c}}^\top {\varvec{\phi }}}}{\partial {\hat{y}}}\right) ^2\right) |\text{ det }B_{a,\theta ,h}|d{\hat{x}}d{\hat{y}}\\&= \varvec{c}^\top M_1\varvec{c}. \end{aligned}$$

By the symbolic computation of Matlab, $M_1$ has $25$ nonzero elements, which read

$$\begin{aligned} M_1(2,2)&= (ah^2\sin \theta )/2,M_1(2,3)=(-ah^2\cos \theta )/2,\\ M_1(2,4)&= (2ah^3 - 2a^2h^3\cos ^3\theta - 2ah^3\cos ^2\theta + 2a^2h^3\cos \theta )/(3\sin \theta ), \\ M_1(2,5)&= -(ah^3(\cos \theta - a + 2a\cos ^2\theta ))/3, M_1(2,6)=-(a^2h^3\sin (2\theta ))/3,\\ M_1(3,2)\!&= \!(ah^2\cos \theta )/2,M_1(3,3)\!=\!(ah^2\sin \theta )/2, M_1(3,4)\!=\!(2ah^3\cos \theta (a\cos \theta \!+\! 1))/3,\\ M_1(3,5)&= (ah^3(\sin \theta + a\sin (2\theta )))/3, M_1(3,6)=(2a^2h^3\sin ^2\theta )/3,\\ M_1(4,2)&= (2ah^3 - 2a^2h^3\cos ^3\theta - 2ah^3\cos ^2\theta + 2a^2h^3\cos \theta )/(3\sin \theta ),\\ M_1(4,3)&= -(2ah^3\cos \theta (a\cos \theta + 1))/3,\\ M_1(4,4)&= (3ah^4 - 5a^2h^4\cos ^3\theta + 3a^3h^4\cos ^2\theta - 3a^3h^4\cos ^4\theta \\&\quad - 3ah^4\cos ^2\theta + 5a^2h^4\cos \theta )/(3\sin \theta )\\ M_1(4,5)&= -(ah^4(6\cos \theta - 5a + 12a^2\cos ^3\theta + 15a\cos ^2\theta - 6a^2\cos \theta ))/12,\\ M_1(4,6)&= -(a^2h^4\sin (2\theta )(6a\cos \theta + 5))/12, M_1(5,2)=(ah^3(a + \cos \theta ))/3,\\ M_1(5,3)&= (ah^3\sin \theta )/3, M_1(5,4)=(ah^4(5a + 6\cos \theta + 5a\cos ^2\theta + 6a^2\cos \theta ))/12,\\ M_1(5,5)&= (ah^4(6(\sin \theta ) a^2 + 5\sin (2\theta )a + 6\sin \theta ))/24, M_1(5,6)=(5a^2h^4\sin ^2\theta )/12\\ M_1(6,2)&= (a^2h^3\sin (2\theta ))/3, M_1(6,3)=(2a^2h^3\sin ^2\theta )/3,\\ M_1(6,4)&= (a^2h^4\sin (2\theta )(6a\cos \theta + 5))/12,\\ M_1(6,5)&= -(a^2h^4(\cos ^2\theta ) - 1)(12a\cos \theta + 5))/12, M_1(6,6)=a^3h^4\sin ^3\theta . \end{aligned}$$

Let Ker($M_2$) and Ker$(M_3)$ be kernel spaces of $M_2$ and $M_3$, respectively. Since

$$\begin{aligned} v^TM_1v\ge 0 \text{ for } \text{ any } v\in \text{ Ker }(M_2)+\text{ Ker }(M_3), \end{aligned}$$

$C_1$ is the smallest generalized eigenvalue of $M_1$ with respect to $M_2$ in the orthogonal complement space Ker($M_2)^\perp$, $C_2$ is the smallest generalized eigenvalue of $M_1$ with respect to $M_3$ in the orthogonal complement space Ker($M_3)^\perp$. Since ranks of both $M_2$ and $M_3$ are $1$, we only need to find nonzero vectors $\varvec{v}_1\in \text{ Ker }(M_2)^\perp$ and $\varvec{v}_2\in \text{ Ker }(M_3)^\perp$, which are eigenvectors corresponding to nonzero eigenvalues of $M_2$ and $M_3$, respectively. This leads to

$$\begin{aligned} C_1=\frac{\varvec{v}^\top _1M_1\varvec{v}_1}{\varvec{v}^\top _1M_2\varvec{v}_1}, C_2=\frac{\varvec{v}^\top _2M_1\varvec{v}_2}{\varvec{v}^\top _2M_3\varvec{v}_2}. \end{aligned}$$

The symbolic computation of Matlab produces that the numerator of $C_1C_2$ is

$$\begin{aligned}&(-36a^6)\cos ^6\theta + (216a^7 + 108a^5)\cos ^5\theta + (252a^8 + 324a^6 - 9a^4)\cos ^4\theta \\&\qquad + (- 324a^7 - 54a^5 + 162a^3)\cos ^3\theta + (- 252a^8 - 117a^6 + 495a^4 + 171a^2)\cos ^2\theta \\&\qquad +(108a^7 - 54a^5 - 162a^3)\cos \theta + 144a^8 + 117a^6 - 54a^4 + 117a^2 + 144, \end{aligned}$$

and that the denominator of $C_1C_2$ is

$$\begin{aligned}&(16a^8)\cos ^8\theta + (-32a^7)\cos ^7\theta + (- 32a^8 + 104a^6)\cos ^6\theta + (-152a^5)\cos ^5\theta \\&\qquad + (48a^8 - 48a^6 + 217a^4)\cos ^4\theta + (- 32a^5 - 208a^3)\cos ^3\theta \!+\! (- 32a^8 + 96a^6 + 54a^4\\&\qquad +152a^2)\cos ^2\theta \!+\! (\!-\! 32a^7 \!-\! 72a^5 \!-\! 48a^3 \!-\! 64a)\cos \theta \!+\! 16a^8 \!+\! 8a^6 + 33a^4 + 8a^2 + 16. \end{aligned}$$