Superconvergence patch recovery for the gradient of the tensor-product linear triangular prism element

Abstract

In this article, we study superconvergence of the finite element approximation to the solution of a general second-order elliptic boundary value problem in three dimensions over a fully uniform mesh of piecewise tensor-product linear triangular prism elements. First, we give the superclose property of the gradient between the finite element solution $u_h$ and the interpolant Πu. Second, we introduce a superconvergence recovery scheme for the gradient of the finite element solution. Finally, superconvergence of the recovered gradient is derived.

1 Introduction

Superconvergence of the gradient for the finite element approximation is a phenomenon whereby the convergent order of the derivatives of the finite element solutions exceeds the optimal global rate. Up to now, superconvergence is still an active research topic (see [1–6]). Recently, we studied the superconvergence patch recovery (SPR) technique introduced by Zienkiewicz and Zhu [7–9] for the linear tetrahedral element and proved pointwise superconvergent property of the recovered gradient by SPR. For the linear tetrahedral element, Chen and Wang [10] also discussed superconvergent properties of the gradients by SPR and obtained superconvergence results of the recovered gradients in the average sense of the $L^2$-norm. In addition, Chen [11] and Goodsell [12] derived superconvergence estimates of the recovered gradient by the $L^2$-projection technique and the average technique, respectively. This article will use the SPR technique to obtain a superconvergence estimate for the gradient of the tensor-product linear triangular prism element. In this article, we shall use the letter C to denote a generic constant which may not be the same in each occurrence and also use the standard notations for the Sobolev spaces and their norms.

2 General elliptic boundary value problem and finite element discretization

We consider the model problem

$$\mathcal{L}u\equiv -\sum _{i,j=1}^3{\partial }_j(a_{ij}{\partial }_{i}u)+\sum _{i=1}^3{a}_i{\partial }_{i}u+a_{0}u=f\text{in }\mathrm{\Omega },u=0\text{on }\partial \mathrm{\Omega }.$$

(2.1)

Here $\mathrm{\Omega }={[0,1]}^2\times [0,1]={\mathrm{\Omega }}_{xy}\times {\mathrm{\Omega }}_z\subset {\mathcal{R}}^3$ is a rectangular block with boundary ∂ Ω consisting of faces parallel to the x-, y-, and z-axes. We also assume that the given functions $a_{ij},a_i\in W^{1,\mathrm{\infty }}(\mathrm{\Omega })$, $a_0\in L^{\mathrm{\infty }}(\mathrm{\Omega })$, and $f\in L^2(\mathrm{\Omega })$. In addition, we write ${\partial }_{1}u=\frac{\partial u}{\partial x}$, ${\partial }_{2}u=\frac{\partial u}{\partial y}$, and ${\partial }_{3}u=\frac{\partial u}{\partial z}$, which are usual partial derivatives.

To discretize the problem, one proceeds as follows. The domain Ω is firstly partitioned into subcubes of side h, and each of these is then subdivided into two triangular prisms. We denote by $\{{\mathcal{T}}^h\}$ these uniform partitions as above. Thus $\overline{\mathrm{\Omega }}={\bigcup }_{e\in {\mathcal{T}}^h}\overline{e}$. Obviously, we can write $e=D\times L$ (see Figure 1), where D and L represent a triangle parallel to the xy-plane and a one-dimensional interval parallel to the z-axes, respectively.

Figure 1

Triangular prisms partition. This figure gives how to partition the domain Ω. The domain Ω is firstly partitioned into subcubes of side h, and each of these is then subdivided into two triangular prisms.

We introduce a tensor-product linear polynomial space denoted by , that is,

$$q(x,y,z)=\sum _{(i,j,k)\in \mathcal{I}}{a}_{ijk}{x}^i{y}^j{z}^k,a_{ijk}\in \mathcal{R},q\in \mathcal{P},$$

where $\mathcal{P}={\mathcal{P}}_{xy}\otimes {\mathcal{P}}_z$, ${\mathcal{P}}_{xy}$ stands for the linear polynomial space with respect to $(x,y)$, and ${\mathcal{P}}_z$ is the linear polynomial space with respect to z. The indexing set ℐ satisfies $\mathcal{I}=\{(i,j,k)|i,j,k\ge 0,i+j\le 1,k\le 1\}$. Let ${\mathrm{\Pi }}_{xy}^e$ be the linear interpolation operator with respect to $(x,y)\in D$, and let ${\mathrm{\Pi }}_z^e$ be the linear interpolation operator with respect to $z\in L$. Thus we may define the tensor-product linear interpolation operator by ${\mathrm{\Pi }}^e:H_0^1(e)\to \mathcal{P}(e)$. Obviously, ${\mathrm{\Pi }}^e={\mathrm{\Pi }}_{xy}^e\otimes {\mathrm{\Pi }}_z^e={\mathrm{\Pi }}_z^e\otimes {\mathrm{\Pi }}_{xy}^e$. In addition, the weak form of problem (2.1) is

$$a(u,v)=(f,v)\mathrm{\forall }v\in H_0^1(\mathrm{\Omega }),$$

(2.2)

where

$$a(u,v)\equiv {\int }_{\mathrm{\Omega }}(\sum _{i,j=1}^3{a}_{ij}{\partial }_{i}u{\partial }_{j}v+\sum _{i=1}^3{a}_i{\partial }_{i}uv+a_{0}uv)dxdydz,$$

and

$$(f,v)={\int }_{\mathrm{\Omega }}fvdxdydz.$$

Define the tensor-product linear triangular prism finite element space by

$$S_0^h(\mathrm{\Omega })=\{v\in H_0^1(\mathrm{\Omega }):v{|}_e\in \mathcal{P}(e)\mathrm{\forall }e\in {\mathcal{T}}^h\}.$$

Thus, the finite element method of problem (2.2) is to find $u_h\in S_0^h(\mathrm{\Omega })$ such that

$$a(u_h,v)=(f,v)\mathrm{\forall }v\in S_0^h(\mathrm{\Omega }).$$

Moreover, from the definitions of ${\mathrm{\Pi }}^e$ and $S_0^h(\mathrm{\Omega })$, we may define a global tensor-product linear interpolation operator $\mathrm{\Pi }:H_0^1(\mathrm{\Omega })\to S_0^h(\mathrm{\Omega })$. Here $(\mathrm{\Pi }u){|}_e={\mathrm{\Pi }}^{e}u$. As for this interpolation operator, the following Lemma 2.1 holds (see [13]).

Lemma 2.1 For Πu and $u_h$, the tensor-product linear interpolant and the tensor-product linear triangular prism finite element approximation to u, respectively, we have the supercloseness estimate

$$|u_h-\mathrm{\Pi }u{|}_{1,\mathrm{\infty },\mathrm{\Omega }}\le C{h}^2{|lnh|}^{\frac{4}{3}}{\parallel u\parallel }_{3,\mathrm{\infty },\mathrm{\Omega }}.$$

(2.3)

3 Gradient recovery and superconvergence

For $v\in S_0^h(\mathrm{\Omega })$, we consider a SPR scheme of ∇v. We denote by $R_h$ the SPR-recovery operator for ∇v and begin by defining the point values of $R_{h}v$ at the element nodes. After the recovered values at all nodes are obtained, we construct a tensor-product linear interpolation by using these values, namely SPR-recovery gradient $R_{h}v$. Obviously, $R_{h}v\in {(S_0^h(\mathrm{\Omega }))}^3$.

Let us first assume that N is an interior node of the partition ${\mathcal{T}}^h$, and denote by ω the element patch around N containing 12 triangular prisms. Under the local coordinate system centered N, we let $Q_i(x_i,y_i,z_i)$ be the barycenter of a triangular prism $e_i\subset \omega$, $i=1,2,\dots ,12$. Obviously, ${\sum }_{i=1}^{12}(x_i,y_i,z_i)=(0,0,0)$. SPR uses the discrete least-squares fitting to seek linear vector function $\mathbf{p}\in {(P_1(\omega ))}^3$ such that

$$\sum _{i=1}^{12}[\mathbf{p}(Q_i)-\mathrm{\nabla }v(Q_i)]q(Q_i)=(0,0,0)\mathrm{\forall }q\in P_1(\omega ),$$

(3.1)

where $v\in S_0^h(\mathrm{\Omega })$. We define $R_{h}v(N)=\mathbf{p}(0,0,0)$. If N is a boundary node, we calculate $R_{h}v(N)$ by linear extrapolation from the values of $R_{h}v$ already obtained at two neighboring interior nodes, $N_1$ and $N_2$ (with diagonal directions being used for edge nodes and corner nodes) (see Figure 2). Namely,

$$R_{h}v(N)=2R_{h}v(N_1)-R_{h}v(N_2).$$

Figure 2

N , a boundary node. $N_1$ and $N_2$ are interior nodes. We can calculate $R_{h}v(N)$ by linear extrapolation from the values of $R_{h}v$ already obtained at two neighboring interior nodes, $N_1$ and $N_2$.

Lemma 3.1 Let ω be the element patch around an interior node N, and $u\in W^{3,\mathrm{\infty }}(\omega )$. For $\mathrm{\Pi }u\in S_0^h(\mathrm{\Omega })$ the interpolant to u, we have

$$|\mathrm{\nabla }u(N)-R_h\mathrm{\Pi }u(N)|\le C{h}^2{\parallel u\parallel }_{3,\mathrm{\infty },\omega }.$$

(3.2)

Proof Choose $v=\mathrm{\Pi }u$ and set $q=1$ in (3.1) to obtain ${\sum }_{i=1}^{12}\mathbf{p}(Q_i)={\sum }_{i=1}^{12}\mathrm{\nabla }\mathrm{\Pi }u(Q_i)$. Therefore,

$$\begin{array}{rl}{R}_h\mathrm{\Pi }u(N)-\frac{1}{12}\sum _{i=1}^{12}\mathrm{\nabla }\mathrm{\Pi }u(Q_i)& =\mathbf{p}(0,0,0)-\frac{1}{12}\sum _{i=1}^{12}\mathbf{p}(x_i,y_i,z_i)\\ =-\frac{1}{12}\sum _{i=1}^{12}[{\partial }_1\mathbf{p}(0,0,0)x_i+{\partial }_2\mathbf{p}(0,0,0)y_i+{\partial }_3\mathbf{p}(0,0,0)z_i]\\ =(0,0,0).\end{array}$$

That is,

$$R_h\mathrm{\Pi }u(N)=\frac{1}{12}\sum _{i=1}^{12}\mathrm{\nabla }\mathrm{\Pi }u(Q_i).$$

(3.3)

Further,

$$\begin{array}{r}\frac{1}{12}\sum _{i=1}^{12}\mathrm{\nabla }\mathrm{\Pi }u(Q_i)-\mathrm{\nabla }u(N)\\ =\frac{1}{12}\sum _{i=1}^{12}\mathrm{\nabla }(\mathrm{\Pi }u-u)(Q_i)+\frac{1}{12}\sum _{i=1}^{12}[\mathrm{\nabla }u(Q_i)-\mathrm{\nabla }u(N)]\\ =\frac{1}{12}\sum _{i=1}^{12}\mathrm{\nabla }(\mathrm{\Pi }u-u)(Q_i)\\ +\frac{1}{12}\sum _{i=1}^{12}[{\partial }_1\mathrm{\nabla }u(N)x_i+{\partial }_2\mathrm{\nabla }u(N)y_i+{\partial }_3\mathrm{\nabla }u(N)z_i]+\mathbf{r}(u),\end{array}$$

(3.4)

where $\frac{1}{12}{\sum }_{i=1}^{12}[{\partial }_1\mathrm{\nabla }u(N)x_i+{\partial }_2\mathrm{\nabla }u(N)y_i+{\partial }_3\mathrm{\nabla }u(N)z_i]=(0,0,0)$, and the high-order term $\mathbf{r}(u)$ satisfies $|\mathbf{r}(u)|\le C{h}^2{|u|}_{3,\mathrm{\infty },\omega }$.

In (3.4), we write $\mathbf{f}(u)=\frac{1}{12}{\sum }_{i=1}^{12}\mathrm{\nabla }(\mathrm{\Pi }u-u)(Q_i)$. For every $u\in P_2(\omega )$, it is not difficult to verify $\mathbf{f}(u)=(0,0,0)$. Thus, by the Bramble-Hilbert lemma [14], we have $|\mathbf{f}(u)|\le C{h}^2{|u|}_{3,\mathrm{\infty },\omega }$. Therefore,

$$|\frac{1}{12}\sum _{i=1}^{12}\mathrm{\nabla }\mathrm{\Pi }u(Q_i)-\mathrm{\nabla }u(N)|\le C{h}^2{|u|}_{3,\mathrm{\infty },\omega }.$$

(3.5)

Combining (3.3) and (3.5), we obtain the result (3.2). □

Lemma 3.2 For $\mathrm{\Pi }u\in S_0^h(\mathrm{\Omega })$ the tensor-product linear interpolant to u, the solution of (2.2), and $R_h$ the SPR recovery operator, we have the superconvergent estimate

$$|\mathrm{\nabla }u-R_h\mathrm{\Pi }u{|}_{0,\mathrm{\infty },\mathrm{\Omega }}\le C{h}^2{\parallel u\parallel }_{3,\mathrm{\infty },\mathrm{\Omega }}.$$

(3.6)

Proof Denote by $F:\hat{e}\to e$ an affine transformation. Obviously, there exists an element $e\in {\mathcal{T}}^h$, using the triangle inequality and the Sobolev embedding theorem [15], and (3.3), such that

$$\begin{array}{rcl}|\mathrm{\nabla }u-R_h\mathrm{\Pi }u{|}_{0,\mathrm{\infty },\mathrm{\Omega }}& =& |\mathrm{\nabla }u-R_h\mathrm{\Pi }u{|}_{0,\mathrm{\infty },e}\\ \le & C{h}^{-1}|\mathrm{\nabla }\hat{u}-R_h\widehat{\mathrm{\Pi }u}{|}_{0,\mathrm{\infty },\hat{e}}\\ \le & C{h}^{-1}[|\mathrm{\nabla }\hat{u}{|}_{0,\mathrm{\infty },\hat{e}}+|R_h\widehat{\mathrm{\Pi }u}{|}_{0,\mathrm{\infty },\hat{e}}]\\ \le & C{h}^{-1}[|\mathrm{\nabla }\hat{u}{|}_{0,\mathrm{\infty },\hat{\chi }}+|\widehat{\mathrm{\Pi }u}{|}_{1,\mathrm{\infty },\hat{\chi }}]\\ \le & C{h}^{-1}{\parallel \hat{u}\parallel }_{3,\mathrm{\infty },\hat{\chi }},\end{array}$$

where $\hat{\chi }$ is a small patch of elements surrounding the triangular prism, $\hat{e}$. Due to the fact that for $\hat{u}$ quadratic over $\hat{\chi }$,

$$\mathrm{\nabla }\hat{u}-R_h\widehat{\mathrm{\Pi }u}=(0,0,0)\text{in }\hat{\chi },$$

so, from the Bramble-Hilbert lemma [14],

$$|\mathrm{\nabla }u-R_h\mathrm{\Pi }u{|}_{0,\mathrm{\infty },\mathrm{\Omega }}\le C{h}^{-1}|\hat{u}{|}_{3,\mathrm{\infty },\hat{\chi }}\le C{h}^2|u{|}_{3,\mathrm{\infty },\mathrm{\Omega }},$$

which completes the proof of the result (3.6). Finally, we give the main result in this article. □

Theorem 3.1 For $u_h\in S_0^h(\mathrm{\Omega })$ the tensor-product linear triangular prism finite element approximation to u, the solution of (2.2), and $R_h$ the SPR recovery operator, we have the superconvergence estimate

$$|\mathrm{\nabla }u-R_h{u}_h{|}_{0,\mathrm{\infty },\mathrm{\Omega }}\le C{h}^2{|lnh|}^{\frac{4}{3}}{\parallel u\parallel }_{3,\mathrm{\infty },\mathrm{\Omega }}.$$

(3.7)

Proof Using the triangle inequality, we have

$$\begin{array}{rcl}|\mathrm{\nabla }u-R_h{u}_h{|}_{0,\mathrm{\infty },\mathrm{\Omega }}& \le & |R_h(u_h-\mathrm{\Pi }u){|}_{0,\mathrm{\infty },\mathrm{\Omega }}\\ +|\mathrm{\nabla }u-R_h\mathrm{\Pi }u{|}_{0,\mathrm{\infty },\mathrm{\Omega }}\\ \le & |u_h-\mathrm{\Pi }u{|}_{1,\mathrm{\infty },\mathrm{\Omega }}\\ +|\mathrm{\nabla }u-R_h\mathrm{\Pi }u{|}_{0,\mathrm{\infty },\mathrm{\Omega }},\end{array}$$

which combined with (2.3) and (3.6) completes the proof of the result (3.7). □