Regular decomposition and a framework of order reduced methods for fourth order problems

Abstract

This paper is devoted to the construction of order reduced method of fourth order problems. A constructive framework is presented such that a problem on the high-regularity space can be deduced to an equivalent system on three low-regularity spaces which are connected by a regular decomposition corresponding to a decomposition of the regularity of the high order space. The generated numerical schemes based on the deduced problems can be of lower complicacy, and the framework is fit for various fourth order problems. Three fourth order problems are then discussed under the framework, including one in two dimension and two in three dimension. They are each corresponding to a regular decomposition, and thus are discretised based on the discretised analogues of the regular decomposition; optimal error estimates are given.

Appendix A: Detailed calculation of (17)

In this section, we present a detailed proof of the lemma below.

Lemma 19

For a tetrahedron K, there exists a constant C depending on the regularity of K, such that

$$\begin{aligned} \Vert \nabla (p+q)\Vert _{0,K}^2\geqslant C_K(\Vert \nabla p\Vert _{0,K}^2+h_k^{-2}\Vert q\Vert _{0,K}^{2}), \end{aligned}$$

(47)

for any $p\in P_1(K)$ and $q\in P^e(K)$ or $q\in P^f(K)$.

Proof

Denote by $\{\phi _i\}_{i=1}^{10}$ a set of basis functions of $P_1^{+e}(K):=P_1(K)+P^e(K)$, such that $\phi _i=\lambda _i$, $i=1:4$, and $\{\phi _j\}_{j=5}^{10}=\{\lambda _i\lambda _k\}_{1\leqslant i\ne k\leqslant 4}$. Let $A=(a_{ij})_{10\times 10}$ be the element-wise stiffness matrix of $P_1^{+e}(K)$ with respect to $(\nabla \cdot ,\nabla \cdot )$. It is well known that the smallest eigenvalue of A is zero, and there are two positive constants $c_s$ and $C_b$ depending on K only, such that $c_s\leqslant \lambda \leqslant C_b$ for any positive eigenvalue $\lambda$ of A.

We rewrite A in a block formulation:

$$\begin{aligned} A=\left[ \begin{array}{cc} a_{11}&{}\quad A_{RT} \\ A_{LB} &{}\quad A_{RB} \end{array} \right] , \quad A_{RT}=A_{LB}^t\in \mathbb {R}^{1\times 9},\ A_{RB}\in \mathbb {R}^{9\times 9}. \end{aligned}$$

Then, as constant forms the kernel space of $\nabla$, direct calculation leads to that

$$\begin{aligned} \left[ \begin{array}{cc} \mathbb {P} &{}\quad \mathbf {0} \\ \mathbf {0}&{}\quad \mathrm {Id}_{6\times 6} \end{array}\right] A \left[ \begin{array}{cc} \mathbb {P}^t &{}\quad \mathbf {0} \\ \mathbf {0}&{}\quad \mathrm {Id}_{6\times 6} \end{array}\right] = \left[ \begin{array}{cc} 0&{}\quad \mathbf {0} \\ \mathbf {0} &{} A_{RB} \end{array} \right] , \end{aligned}$$

and

$$\begin{aligned} A= \left[ \begin{array}{cc} \mathbb {P}^{-1} &{}\quad \mathbf {0} \\ \mathbf {0}&{}\quad \mathrm {Id}_{6\times 6} \end{array}\right] \left[ \begin{array}{cc} 0&{}\quad \mathbf {0} \\ \mathbf {0} &{}\quad A_{RB} \end{array} \right] \left[ \begin{array}{cc} \mathbb {P}^{-t} &{} \mathbf {0} \\ \mathbf {0}&{}\quad \mathrm {Id}_{6\times 6} \end{array}\right] , \end{aligned}$$

where $\mathbb {P}=\left[ \begin{array}{cccc} 1&{}\quad 1&{}\quad 1&{}\quad 1 \\ &{}\quad 1&{}&{} \\ &{}&{}\quad 1&{} \\ &{}&{}&{}\quad 1 \end{array} \right]$, and $\mathbb {P}^{-1}=\left[ \begin{array}{cccc} 1&{}\quad -1&{}\quad -1&{}\quad -1 \\ &{}\quad 1&{}&{} \\ &{}&{}\quad 1&{} \\ &{}&{}&{}\quad 1 \end{array} \right]$. Preliminary algebraic calculation leads to that, there are two positive constants $\hat{c}_s$ and $\hat{C}_b$, such that $\hat{c}_s\leqslant \hat{\lambda }\leqslant \hat{C}_b$ for all eigenvalues $\hat{\lambda }$ of $A_{RB}$.

Now for $p=\sum _{i=1}^4\mu _i\lambda _i$, and $q=\sum _{i=1}^6\nu _i\phi _i$,

$$\begin{aligned}&\Vert \nabla (p+q)\Vert _{0,K}^2=(\mu _1,\ldots ,\mu _4,\nu _1,\ldots ,\nu _6)A(\mu _1,\ldots ,\mu _4,\nu _1,\ldots ,\nu _6)^t\\&\quad =(\mu _1-\mu _2,\mu _1-\mu _3,\mu _1-\mu _4,\nu _1,\ldots ,\nu _6)\\&\quad A_{RB}(\mu _1-\mu _2,\mu _1-\mu _3,\mu _1-\mu _4,\nu _1,\ldots ,\nu _6)^t \\&\quad \left\{ \begin{array}{l} \displaystyle \geqslant \hat{c}_s\left[ (\mu _1-\mu _2)^2+(\mu _1-\mu _3)^2+(\mu _1-\mu _4)^2+\sum _{i=1}^6\nu _i^2\right] , \\ \displaystyle \leqslant \hat{C}_b\left[ (\mu _1-\mu _2)^2+(\mu _1-\mu _3)^2+(\mu _1-\mu _4)^2+\sum _{i=1}^6\nu _i^2\right] . \end{array} \right. \end{aligned}$$

Similarly, there are two positive constants $\tilde{c}_s$ and $\tilde{C}_b$, such that

$$\begin{aligned} \Vert p+q\Vert _{0,K}^2\left\{ \begin{array}{l} \displaystyle \geqslant \tilde{c}_s |K|\left[ \sum _{i=1}^2\mu _i^2+\sum _{i=1}^6\nu _j^2\right] \\ \displaystyle \leqslant \tilde{C}_b |K|\left[ \sum _{i=1}^2\mu _i^2+\sum _{i=1}^6\nu _j^2\right] . \end{array} \right. \end{aligned}$$

Combine the two representations above together completes the proof for $p\in P_1(K)$ and $q\in P^e(K)$. The proof for the case $q\in P^f(K)$ is the same. $\square$