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Superconvergence Analysis of High-Order Rectangular Edge Elements for Time-Harmonic Maxwell’s Equations

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Abstract

In this paper, high-order rectangular edge elements are used to solve the two dimensional time-harmonic Maxwell’s equations. Superconvergence for the Nédélec interpolation at the Gauss points is proved for both the second and third order edge elements. Using this fact, we obtain the superconvergence results for the electric field \(\mathbf {E}\), magnetic field H and \(curl\mathbf {E}\) in the discrete \(l^2\) norm when the Maxwell’s equations are solved by both elements. Extensive numerical results are presented to justify our theoretical analysis.

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Author information

Authors and Affiliations

  1. School of Mathematics, Shandong University, Jinan, 250100, Shandong, People’s Republic of China

    Ming Sun

  2. Department of Mathematical Sciences, University of Nevada, Las Vegas, NV, 89154-4020, USA

    Jichun Li

  3. School of Mathematics and Information Science, North China University of Water Resources and Electric Power, Zhengzhou, 450011, People’s Republic of China

    Peizhen Wang

  4. Beijing Computational Science Research Center, Beijing, 100193, People’s Republic of China

    Zhimin Zhang

  5. Department of Mathematics, Wayne State University, Detroit, MI, 48202, USA

    Zhimin Zhang

Authors
  1. Ming Sun
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  2. Jichun Li
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  3. Peizhen Wang
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  4. Zhimin Zhang
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Corresponding author

Correspondence to Jichun Li.

Additional information

The first and third authors were partially supported by the China Scholarship Council. The first author was partially supported by National Natural Science Foundation of China (NSFC) Grant 11671233. The second author was partially supported by NSF Grants DMS-1416742, and NSFC 11671340. The fourth author is also supported in part by Grants NSFC 11471031, NSFC 91430216, and NSAF U1530401, and the US NSF Grant DMS-1419040.

Appendices

Appendix 1: The Degrees of Freedom and Basis Functions When k=2

The degrees of freedom \(c_i\) are [18, p.75]:

$$\begin{aligned}&c_i=\int _{l_i}\mathbf {E}(x,y)\cdot \mathbf {\tau }_idl,\ \ \ \ i=1,\ldots ,4,\\&c_5=\int _{l_1}(x-x_c)\mathbf {E}(x,y)\cdot \mathbf {\tau }_1dl,\quad \quad \quad c_6=\int _{l_2}(y-y_c)\mathbf {E}(x,y)\cdot \mathbf {\tau }_2dl,\\&c_7=\int _{l_3}(x-x_c)\mathbf {E}(x,y)\cdot \mathbf {\tau }_3dl,\quad \quad \quad c_8=\int _{l_4}(y-y_c)\mathbf {E}(x,y)\cdot \mathbf {\tau }_4dl,\\&c_9=\int _{K}\mathbf {E}(x,y)\cdot (1,0)'dxdy,\quad \quad \quad c_{10}=\int _{K}\mathbf {E}(x,y)\cdot (0,1)'dxdy,\\&c_{11}=\int _{K}\mathbf {E}(x,y)\cdot (x-x_c,0)'dxdy,\ \ c_{12}=\int _{K}\mathbf {E}(x,y)\cdot (0,y-y_c)'dxdy, \end{aligned}$$

and the basis function \(\varvec{N}_i\) can be expressed as:

$$\begin{aligned}&\varvec{N}_1=\left( \begin{array}{c} \frac{[3(y-y_c)+h_y][(y-y_c)-h_y]}{8h_xh_y^2}\\ 0 \end{array}\right) , \quad \varvec{N}_2=\left( \begin{array}{c} 0\\ \frac{[3(x-x_c)-h_x][(x-x_c)+h_x]}{8h^2_xh_y} \end{array}\right) ,\\&\varvec{N}_3=\left( \begin{array}{c} \frac{[3(y-y_c)-h_y][(y-y_c)+h_y]}{8h_xh_y^2}\\ 0 \end{array}\right) , \quad \varvec{N}_4=\left( \begin{array}{c} 0\\ \frac{[3(x-x_c)+h_x][(x-x_c)-h_x]}{8h^2_xh_y} \end{array}\right) ,\\&\varvec{N}_5=\left( \begin{array}{c} \frac{(x-x_c)[3(y-y_c)-3h_y][3(y-y_c)+h_y]}{8h^3_xh_y^2}\\ 0 \end{array}\right) , \quad \varvec{N}_6=\left( \begin{array}{c} 0\\ \frac{(y-y_c)[3(x-x_c)+3h_x][3(x-x_c)-h_x]}{8h^2_xh^3_y} \end{array}\right) ,\\&\varvec{N}_7=\left( \begin{array}{c} \frac{(x-x_c)[3(y-y_c)+3h_y][3(y-y_c)-h_y]}{8h^3_xh_y^2}\\ 0 \end{array}\right) , \quad \varvec{N}_8=\left( \begin{array}{c} 0\\ \frac{(y-y_c)[3(x-x_c)-3h_x][3(x-x_c)+h_x]}{8h^2_xh^3_y} \end{array}\right) ,\\&\varvec{N}_9=\left( \begin{array}{c} \frac{-3[(y-y_c)-h_y][(y-y_c)+h_y]}{8h_xh_y^3}\\ 0 \end{array}\right) , \quad \varvec{N}_{10}=\left( \begin{array}{c} 0\\ \frac{-3[(x-x_c)-h_x][(x-x_c)+h_x]}{8h^3_xh_y} \end{array}\right) ,\\&\varvec{N}_{11}=\left( \begin{array}{c} \frac{-9(x-x_c)[(y-y_c)-h_y][(y-y_c)+h_y]}{8h^3_xh_y^3}\\ 0 \end{array}\right) , \quad \varvec{N}_{12}=\left( \begin{array}{c} 0\\ \frac{-9(y-y_c)[(x-x_c)+h_x][(x-x_c)-h_x]}{8h^3_xh^3_y} \end{array}\right) . \end{aligned}$$

For the interpolation \(I_{K} H\in Q_{1,1}\), the degrees of freedom \(d_i\) and the basis functions \(\psi _i\) are:

$$\begin{aligned}&d_1=\int _{K}H(x,y)dxdy, \quad \quad \quad \quad \quad d_2=\int _{K}H(x,y)(x-x_c)dxdy,\\&d_3=\int _{K}H(x,y)(y-y_c)dxdy, \quad \ d_4=\int _{K}H(x,y)(x-x_c)(y-y_c)dxdy, \end{aligned}$$

and the basis functions \(\psi _i\) are:

$$\begin{aligned}&\psi _1(x,y)=\frac{1}{4h_xh_y},\quad \quad \quad \psi _2(x,y)=\frac{3(x-x_c)}{4h^3_xh_y},\\&\psi _3(x,y)=\frac{3(y-y_c)}{4h_xh^3_y}, \quad \ \psi _4(x,y)=\frac{9(x-x_c)(y-y_c)}{4h^3_xh^3_y}. \end{aligned}$$

Appendix 2: The Degrees of Freedom and Basis Functions When k=3

The degrees of freedom \(c_i\) are

$$\begin{aligned}&c_i=\int _{l_i}\mathbf {E}(x,y)\cdot \mathbf {\tau }_idl,\ \ \ \ i=1,\ldots ,4,\\&c_5=\int _{l_1}(x-x_c)\mathbf {E}(x,y)\cdot \mathbf {\tau }_1dl,\quad \quad \quad c_6=\int _{l_2}(y-y_c)\mathbf {E}(x,y)\cdot \mathbf {\tau }_2dl,\\&c_7=\int _{l_3}(x-x_c)\mathbf {E}(x,y)\cdot \mathbf {\tau }_3dl,\quad \quad \quad c_8=\int _{l_4}(y-y_c)\mathbf {E}(x,y)\cdot \mathbf {\tau }_4dl,\\&c_9=\int _{l_1}(x-x_c)^2\mathbf {E}(x,y)\cdot \mathbf {\tau }_1dl,\quad \quad \quad c_{10}=\int _{l_2}(y-y_c)^2\mathbf {E}(x,y)\cdot \mathbf {\tau }_2dl,\\&c_{11}=\int _{l_3}(x-x_c)^2\mathbf {E}(x,y)\cdot \mathbf {\tau }_3dl,\quad \quad \ \ c_{12}=\int _{l_4}(y-y_c)^2\mathbf {E}(x,y)\cdot \mathbf {\tau }_4dl,\\&c_{13}=\int _{K}\mathbf {E}(x,y)\cdot (1,0)'dxdy,\quad \quad \quad \ c_{14}=\int _{K}\mathbf {E}(x,y)\cdot (0,1)'dxdy,\\&c_{15}=\int _{K}\mathbf {E}(x,y)\cdot (x-x_c,0)'dxdy,\quad \ c_{16}=\int _{K}\mathbf {E}(x,y)\cdot (0,y-y_c)'dxdy,\\&c_{17}=\int _{K}\mathbf {E}(x,y)\cdot (y-y_c,0)'dxdy,\quad \ c_{18}=\int _{K}\mathbf {E}(x,y)\cdot (0,x-x_c)'dxdy,\\&c_{19}=\int _{K}\mathbf {E}(x,y)\cdot \big ((x-x_c)(y-y_c),0\big )'dxdy,\\&c_{20}=\int _{K}\mathbf {E}(x,y)\cdot \big (0,(x-x_c)(y-y_c)\big )'dxdy,\\&c_{21}=\int _{K}\mathbf {E}(x,y)\cdot \big ((x-x_c)^2,0\big ) 'dxdy,\quad \quad \quad \ c_{22}=\int _{K}\mathbf {E}(x,y)\cdot \big (0,(y-y_c)^2\big )'dxdy,\\&c_{23}=\int _{K}\mathbf {E}(x,y)\cdot \big ((x-x_c)^2(y-y_c),0\big )'dxdy,\\&c_{24}=\int _{K}\mathbf {E}(x,y)\cdot \big (0,(x-x_c)(y-y_c)^2\big )'dxdy, \end{aligned}$$

and the basis function \(\varvec{N}_i\) can be expressed as:

$$\begin{aligned}&\varvec{N}_1=\left( \begin{array}{c} \frac{3[5(x-x_c)^2-3h_x^2][6(y-y_c)^2-(y-y_c-h_y)^2][(y-y_c)-h_y]}{32h^3_xh_y^3}\\ 0 \end{array}\right) ,\\&\varvec{N}_2=\left( \begin{array}{c} 0\\ \frac{-3[5(y-y_c)^2-3h_y^2][6(x-x_c)^2-(x-x_c+h_x)^2][(x-x_c)+h_x]}{32h^3_xh^3_y} \end{array}\right) ,\\&\varvec{N}_3=\left( \begin{array}{c} \frac{-3[5(x-x_c)^2-3h_x^2][6(y-y_c)^2-(y-y_c+h_y)^2][(y-y_c)+h_y]}{32h^3_xh_y^3}\\ 0 \end{array}\right) ,\\&\varvec{N}_4=\left( \begin{array}{c} 0\\ \frac{3[5(y-y_c)^2-3h_y^2][6(x-x_c)^2-(x-x_c-h_x)^2][(x-x_c)-h_x]}{32h^3_xh^3_y} \end{array}\right) , \end{aligned}$$
$$\begin{aligned}&\varvec{N}_5=\left( \begin{array}{c} \frac{-3(x-x_c)[6(y-y_c)^2-(y-y_c-h_y)^2][(y-y_c)-h_y]}{8h^3_xh_y^3}\\ 0 \end{array}\right) ,\\&\varvec{N}_6=\left( \begin{array}{c} 0\\ \frac{3(y-y_c)[6(x-x_c)^2-(x-x_c+h_x)^2][(x-x_c)+h_x]}{8h^3_xh^3_y} \end{array}\right) ,\\&\varvec{N}_7=\left( \begin{array}{c} \frac{3(x-x_c)[6(y-y_c)^2-(y-y_c+h_y)^2][(y-y_c)+h_y]}{8h^3_xh_y^3}\\ 0 \end{array}\right) ,\\&\varvec{N}_8=\left( \begin{array}{c} 0\\ \frac{-3(y-y_c)[6(x-x_c)^2-(x-x_c-h_x)^2][(x-x_c)-h_x]}{8h^3_xh^3_y} \end{array}\right) ,\\&\varvec{N}_9=\left( \begin{array}{c} \frac{-15[3(x-x_c)^2-h_x^2][6(y-y_c)^2-(y-y_c-h_y)^2][(y-y_c)-h_y]}{32h^5_xh_y^3}\\ 0 \end{array}\right) ,\\&\varvec{N}_{10}=\left( \begin{array}{c} 0\\ \frac{15[3(y-y_c)^2-h_y^2][6(x-x_c)^2-(x-x_c+h_x)^2][(x-x_c)+h_x]}{32h^3_xh^5_y} \end{array}\right) ,\\&\varvec{N}_{11}=\left( \begin{array}{c} \frac{15[3(x-x_c)^2-h_x^2][6(y-y_c)^2-(y-y_c+h_y)^2][(y-y_c)+h_y]}{32h^5_xh_y^3}\\ 0 \end{array}\right) ,\\&\varvec{N}_{12}=\left( \begin{array}{c} 0\\ \frac{-15[3(y-y_c)^2-h_y^2][6(x-x_c)^2-(x-x_c-h_x)^2][(x-x_c)-h_x]}{32h^3_xh^5_y} \end{array}\right) ,\\ \end{aligned}$$
$$\begin{aligned}&\varvec{N}_{13}=\left( \begin{array}{c} \frac{9[5(x-x_c)^2-3h_x^2][(y-y_c)^2-h_y^2]}{32h^3_xh_y^3}\\ 0 \end{array}\right) , \quad \varvec{N}_{14}=\left( \begin{array}{c} 0\\ \frac{9[5(y-y_c)^2-3h_y^2][(x-x_c)^2-h_x^2]}{32h^3_xh^3_y} \end{array}\right) ,\\&\varvec{N}_{15}=\left( \begin{array}{c} \frac{-9(x-x_c)[(y-y_c)^2-h_y^2]}{8h^3_xh_y^3}\\ 0 \end{array}\right) , \quad \varvec{N}_{16}=\left( \begin{array}{c} 0\\ \frac{-9(y-y_c)[(x-x_c)^2-h_x^2]}{8h^3_xh^3_y} \end{array}\right) ,\\&\varvec{N}_{17}=\left( \begin{array}{c} \frac{45[5(x-x_c)^2-3h_x^2][(y-y_c)^2-h_y^2](y-y_c)}{32h^3_xh_y^5}\\ 0 \end{array}\right) , \quad \varvec{N}_{18}=\left( \begin{array}{c} 0\\ \frac{45[5(y-y_c)^2-3h_y^2][(x-x_c)^2-h_x^2](x-x_c)}{32h^5_xh^3_y} \end{array}\right) ,\\&\varvec{N}_{19}=\left( \begin{array}{c} \frac{-45(x-x_c)[(y-y_c)^2-h_y^2](y-y_c)}{8h^3_xh_y^5}\\ 0 \end{array}\right) , \quad \varvec{N}_{20}=\left( \begin{array}{c} 0\\ \frac{-45(y-y_c)[(x-x_c)^2-h_x^2](x-x_c)}{8h^5_xh^3_y} \end{array}\right) ,\\&\varvec{N}_{21}=\left( \begin{array}{c} \frac{-45[3(x-x_c)^2-h_x^2][(y-y_c)^2-h_y^2]}{32h^5_xh_y^3}\\ 0 \end{array}\right) , \quad \varvec{N}_{22}=\left( \begin{array}{c} 0\\ \frac{-45[3(y-y_c)^2-h_y^2][(x-x_c)^2-h_x^2]}{32h^3_xh^5_y} \end{array}\right) ,\\&\varvec{N}_{23}=\left( \begin{array}{c} \frac{-225[3(x-x_c)^2-h_x^2][(y-y_c)^2-h_y^2](y-y_c)}{32h^5_xh_y^5}\\ 0 \end{array}\right) , \quad \varvec{N}_{24}=\left( \begin{array}{c} 0\\ \frac{-225[3(y-y_c)^2-h_y^2][(x-x_c)^2-h_x^2](x-x_c)}{32h^5_xh^5_y} \end{array}\right) . \end{aligned}$$

For the interpolation \(I_{\hat{K}} H\in Q_{2,2}(\hat{K})\), the degrees of freedom \(d_i\) are:

$$\begin{aligned}&d_1=\int _{\hat{K}}H(\hat{x},\hat{y})d\hat{x}d\hat{y}, \quad \quad d_2=\int _{\hat{K}}H(\hat{x},\hat{y})\hat{x}d\hat{x}d\hat{y}, \quad \ d_3=\int _{\hat{K}}H(\hat{x},\hat{y})\hat{y}d\hat{x}d\hat{y},\\&d_4=\int _{\hat{K}}H(\hat{x},\hat{y})\hat{x}\hat{y}d\hat{x}d\hat{y}, \quad d_5=\int _{\hat{K}}H(\hat{x},\hat{y})\hat{x}^2d\hat{x}d\hat{y}, \quad d_6=\int _{\hat{K}}H(\hat{x},\hat{y})\hat{y}^2d\hat{x}d\hat{y},\\&d_7=\int _{\hat{K}}H(\hat{x},\hat{y})\hat{x}^2\hat{y}d\hat{x}d\hat{y}, \ \ d_8=\int _{\hat{K}}H(\hat{x},\hat{y})\hat{x}\hat{y}^2d\hat{x}d\hat{y}, \ \ d_9=\int _{\hat{K}}H(\hat{x},\hat{y})\hat{x}^2\hat{y}^2d\hat{x}d\hat{y}, \end{aligned}$$

and the basis function \(\hat{\psi }_i\) are:

$$\begin{aligned}&\hat{\psi }_1(\hat{x},\hat{y})=\frac{81}{64}-\frac{135}{64}\hat{x}^2 -\frac{135}{64}\hat{y}^2+\frac{225}{64}\hat{x}^2\hat{y}^2,\\&\hat{\psi }_2(\hat{x},\hat{y})=\frac{27}{16}\hat{x}-\frac{45}{16}\hat{x}\hat{y}^2, \ \ \hat{\psi }_3(\hat{x},\hat{y})=\frac{27}{16}\hat{y}-\frac{45}{16}\hat{x}^2\hat{y}, \ \ \hat{\psi }_4(\hat{x},\hat{y})=\frac{9}{4}\hat{x}\hat{y},\\&\hat{\psi }_5(\hat{x},\hat{y})=-\,\frac{135}{64}+\frac{405}{64}\hat{x}^2+\frac{225}{64}\hat{y}^2-\frac{675}{64}\hat{x}^2\hat{y}^2,\\&\hat{\psi }_6(\hat{x},\hat{y})=-\,\frac{135}{64}+\frac{225}{64}\hat{x}^2+\frac{405}{64}\hat{y}^2-\frac{675}{64}\hat{x}^2\hat{y}^2,\\&\hat{\psi }_7(\hat{x},\hat{y})=-\,\frac{45}{16}\hat{y}+\frac{135}{16}\hat{x}^2\hat{y}, \ \ \hat{\psi }_8(\hat{x},\hat{y})=-\,\frac{45}{16}\hat{x}+\frac{135}{16}\hat{x}\hat{y}^2,\\&\hat{\psi }_9(\hat{x},\hat{y})=\frac{225}{64}-\frac{675}{64}\hat{x}^2-\frac{675}{64}\hat{y}^2+\frac{2025}{64}\hat{x}^2\hat{y}^2. \end{aligned}$$

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Sun, M., Li, J., Wang, P. et al. Superconvergence Analysis of High-Order Rectangular Edge Elements for Time-Harmonic Maxwell’s Equations. J Sci Comput 75, 510–535 (2018). https://doi.org/10.1007/s10915-017-0544-2

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  • DOI: https://doi.org/10.1007/s10915-017-0544-2

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