A High-Order Finite Difference Scheme for 3D Unsteady Convection Diffusion Reaction Equations
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摘要:
针对三维非稳态对流扩散反应方程,构造了一种高精度紧致有限差分格式，对空间的离散采用四阶紧致差分方法，对时间的离散采用Taylor级数展开和余项修正技术，所提格式在时间上的精度为二阶、在空间上的精度为四阶。利用Fourier稳定性分析法证明了该格式是无条件稳定的。最后给出数值算例验证了理论结果。
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关键词:
- 对流扩散反应方程 /
- 变对流项和反应项系数 /
- 高精度紧致格式 /
- 无条件稳定 /
- 有限差分法
Abstract:Based on the 4th-order compact difference scheme for spatial discretization, the Taylor series expansion and the error remainder correction method for temporal discretization, a high-order compact finite difference scheme for solving the 3D unsteady convection diffusion reaction equations was proposed. The unconditional stability was proved with the Fourier analysis method. The proposed scheme has 2nd-order accuracy in time and 4th-order accuracy in space. At last, numerical examples validate the theoretical results.
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图 1 问题2当
$ h = 0.025{\text{,}}t = 1.25{\text{,}}p = q = r = 0.8{\text{,}}\tau = 6.25 \times {10^{ - 3}} $ 时，在区域$ 1.2 \leqslant x,y \leqslant 1.8 $ 内，截面$ z = 1.5 $ 上的等值线图：(a) PHOC-ADI格式和精确解；(b) EHOC-ADI格式和精确解；(c) RHOC-ADI格式和精确解;(d) 本文格式和精确解Figure 1. Numerical contour lines compared with exact solutions at
$ z = 1.5 $ in region$ 1.2 \leqslant x,y \leqslant 1.8 $ in problem 2 for$ h = 0.025,\;t = 1.25,$ $p = q = r = 0.8{\text{,}}\;\tau = 6.25 \times {10^{ - 3}}$ : (a) PHOC-ADI scheme and exact solutions; (b) EHOC-ADI scheme and exact solutions; (c) RHOC-ADI scheme and exact solutions; (d) the present scheme and exact solutions
图 2 问题2当
$h = 0.025,\;t = 1.25 \times {10^{-4}},\;p = q = r = 8\;000,\;\tau = 6.25 \times {10^{ - 7}}$ 时，在区域$ 1.2 \leqslant x,y \leqslant 1.8 $ 内，截面z=1.5上的等值线图：(a) PHOC-ADI格式和精确解；(b) EHOC-ADI格式和精确解；(c) RHOC-ADI格式和精确解；(d) 本文格式和精确解Figure 2. Numerical contour lines compared with exact solutions at
${{z}} = 1.5$ in region$ 1.2 \leqslant x,y \leqslant 1.8 $ in problem 2 for$h = 0.025,\;t = 1.25 \times {10^{-4}},\; $ $ p = q = r = 8\;000,\;\tau = 6.25 \times {10^{ - 7}}$ : (a) PHOC-ADI scheme and exact solutions; (b) EHOC-ADI scheme and exact solutions; (c) RHOC-ADI scheme and exact solutions;(d) the present scheme and exact solutions表 1 问题1当
$\tau = {h^2}, t = 0.25$ 时的$ {L_\infty } $ 误差和$ {L_2} $ 误差及收敛阶Table 1.
$ {L_\infty } $ errors,$ {L_2} $ errors and convergence rates for$\tau = {h^2},\;t = 0.25$ in problem 1$ p,q,r $ $ h $ C-N scheme BTCS scheme present scheme $ {L_\infty } $ $\delta $ $ {L_2} $ $\delta $ $ {L_\infty } $ $\delta $ $ {L_2} $ $\delta $ $ {L_\infty } $ $\delta $ $ {L_2} $ $\delta $ case ① 1/8 1.60E−2 5.62E−3 1.63E−2 5.73E−3 2.74E−4 1.67E−4 1/16 3.97E−3 2.01 1.40E−3 2.01 4.05E−3 2.00 1.42E−3 2.00 7.81E−6 5.13 3.87E−6 5.43 1/32 9.90E−4 2.00 3.48E−4 2.00 1.01E−3 2.00 3.56E−4 2.00 3.94E−7 4.30 1.62E−7 4.57 1/64 2.47E−4 2.00 8.71E−5 2.00 2.53E−4 2.00 8.89E−5 2.00 2.29E−8 4.10 8.08E−9 4.32 case ② 1/8 1.58E−2 5.55E−3 1.62E−2 5.66E−3 2.83E−4 1.70E−4 1/16 3.94E−3 2.00 1.38E−3 2.01 4.02E−3 2.01 1.41E−3 2.01 8.48E−6 5.06 4.10E−6 5.37 1/32 9.83E−4 2.00 3.44E−4 2.00 1.00E−3 2.01 3.52E−4 2.00 4.30E−7 4.30 1.76E−7 4.54 1/64 2.46E−4 2.00 8.61E−5 2.00 2.51E−4 1.99 8.79E−5 2.00 2.52E−8 4.09 9.75E−9 4.17 
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表 2 问题1当
$h = 0.031\;25, t = 0.5$ 时，本文格式的$ {L_\infty } $ 误差、$ {L_2} $ 误差、收敛阶及CPU时间Table 2.
$ {L_\infty } $ errors,$ {L_2} $ errors, convergence rates and CPU time for$h = 0.031\;25,\;t = 0.5$ in problem 1$ p,q,r $ $ \tau $ $ {L_\infty } $ $\delta $ $ {L_2} $ $\delta $ CPU time tCPU/s case ① 0.1 4.883E−4 1.654E−4 195.934 0.05 1.234E−4 1.98 4.076E−5 2.01 275.417 0.025 3.155E−5 1.97 1.050E−5 1.96 334.355 0.0125 8.050E−6 1.97 2.695E−6 1.96 406.406 0.00625 2.296E−6 1.81 7.981E−7 1.76 455.322 case ② 0.1 8.220E−4 2.528E−4 240.216 0.05 2.017E−4 2.03 6.189E−5 2.03 345.248 0.025 5.056E−5 2.00 1.550E−5 2.00 426.347 0.0125 1.292E−5 1.97 3.920E−6 1.98 520.264 0.00625 3.519E−6 1.88 1.039E−6 1.92 539.716
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表 3 问题1当
$h = {1 \mathord{\left/ {\vphantom {1 {32}}} \right. } {32}}, t = 1$ 时，对不同网格比$\lambda = {\tau \mathord{/ {\vphantom {\tau {{h^2}}}} } {{h^2}}}$ 的$ {L_\infty } $ 误差和$ {L_2} $ 误差Table 3.
$ {L_\infty } $ errors and$ {L_2} $ errors for$h = {1 \mathord{\left/ {\vphantom {1 {32}}} \right. } {32}}{\text{,}}\;t = 1$ for different mesh ratio$\lambda = {\tau \mathord{/ {\vphantom {\tau {{h^2}}}} } {{h^2}}}$ values in problem 1$ p,q,r $ $ \lambda $ number of steps in the time direction C-N scheme BTCS scheme present scheme $ {L_\infty } $ $ {L_2} $ $ {L_\infty } $ $ {L_2} $ $ {L_\infty } $ $ {L_2} $ case ① 0.8 1280 2.044E−3 7.086E−4 2.079E−3 7.213E−4 8.703E−7 3.611E−7 1.6 640 2.043E−3 7.086E−4 2.114E−3 7.340E−4 9.715E−7 4.008E−7 3.2 320 2.043E−3 7.085E−4 2.185E−3 7.595E−4 1.495E−6 5.737E−7 6.4 160 2.043E−3 7.083E−4 2.327E−3 8.104E−4 3.841E−6 1.330E−6 case ② 0.8 1280 2.110E−3 7.362E−4 2.144E−3 7.485E−4 7.827E−7 3.002E−7 1.6 640 2.110E−3 7.362E−4 2.178E−3 7.608E−4 7.906E−7 2.496E−7 3.2 320 2.110E−3 7.364E−4 2.246E−3 7.854E−4 1.381E−6 4.394E−7 6.4 160 2.112E−3 7.371E−4 2.383E−3 8.345E−4 6.839E−6 2.142E−6
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表 4 问题2当
$ h = 0.025 $ 时，在不同参数下的$ {L_\infty } $ 误差和$ {L_2} $ 误差Table 4.
$ {L_\infty } $ errors and$ {L_2} $ errors for$ h = 0.025 $ in problem 2scheme L∞ L2 $ t = 1.25,p = q = r = 0.8,\tau = 6.25 \times {10^{ - 3}} $ Douglas-Gunn ADI[1] 4.107E−3 5.674E−4 HOC-ADI[3] 1.454E−4 1.626E−5 PHOC-ADI[4] 1.444E−4 1.546E−5 EHOC-ADI[5] 1.044E−3 5.344E−5 RHOC-ADI[6] 1.039E−3 6.245E−5 present scheme 2.196E−4 3.434E−5 $ t = 1.25 \times {10^{ - 1}},p = q = r = 8,\tau = 6.25 \times {10^{ - 4}} $ Douglas-Gunn ADI[1] 1.963E−1 1.158E−2 HOC-ADI[3] 8.886E−2 3.530E−3 PHOC-ADI[4] 8.885E−2 3.529E−3 EHOC-ADI[5] 4.459E−2 1.531E−3 RHOC-ADI[6] 7.983E−3 3.777E−4 present scheme 1.566E−2 8.319E−4 $ t = 1.25 \times {10^{ - 2}},p = q = r = 80,\tau = 6.25 \times {10^{ - 5}} $ Douglas-Gunn ADI[1] 4.455E−1 2.285E−2 HOC-ADI[3] 3.010E−1 1.223E−2 PHOC-ADI[4] 3.018E−1 1.223E−2 EHOC-ADI[5] 1.375E−1 3.782E−3 RHOC-ADI[6] 2.674E−2 9.888E−4 present scheme 4.833E−2 1.939E−3 $ t = 1.25 \times {10^{ - 3}},p = q = r = 800,\tau = 6.25 \times {10^{ - 6}} $ Douglas-Gunn ADI[1] 4.874E−1 2.470E−2 HOC-ADI[3] 1.178E+20 4.617E+19 PHOC-ADI[4] 3.248E−1 1.411E−2 EHOC-ADI[5] 1.567E−1 4.199E−3 RHOC-ADI[6] 3.137E−2 1.115E−3 present scheme 5.562E−2 2.153E−3 $t = 1.25 \times {10^{ -4} },p = q = r = 8\;000,\tau = 6.25 \times {10^{ - 7} }$ Douglas-Gunn ADI[1] 4.918E−1 2.490E−2 HOC-ADI[3] 2.639E+24 1.523E+24 PHOC-ADI[4] 3.267E−1 1.433E−2 EHOC-ADI[5] 1.588E−1 4.244E−3 RHOC-ADI[6] 3.189E−2 1.129E−3 present scheme 5.642E−2 2.176E−3
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表 5 问题2当
$ t = 0.25,\tau = {h^2},p = q = r = 0.8 $ 时，本文格式的$ {L_\infty } $ 误差和$ {L_2} $ 误差及收敛阶Table 5.
$ {L_\infty } $ errors,$ {L_2} $ errors and convergence rates for$ t = 0.25,\tau = {h^2},p = q = r = 0.8 $ in problem 2$ h $ $ {L_\infty } $ $\delta $ $ {L_2} $ $\delta $ 1/40 2.287E−4 2.861E−5 1/60 4.586E−5 3.96 5.516E−6 4.06 1/80 1.470E−5 3.95 1.730E−6 4.03 1/100 6.013E−6 4.01 7.057E−7 4.02 1/120 2.886E−6 4.03 3.396E−7 4.01
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表 6 问题3当
$\tau = {h^2},\;\alpha = 0.1,\;t = 0.5$ 时的$ {L_\infty } $ 误差和$ {L_2} $ 误差及收敛阶Table 6.
$ {L_\infty } $ errors,$ {L_2} $ errors and convergence rates for$\tau = {h^2},\;\alpha = 0.1,\,t = 0.5$ in problem 3$ h $ DHOC[20] present scheme $ {L_\infty } $ $\delta $ $ {L_2} $ $\delta $ $ {L_\infty } $ $\delta $ $ {L_2} $ $\delta $ 1/8 9.079E−5 3.060E−5 5.848E−5 2.078E−5 1/16 2.555E−6 5.15 9.067E−7 5.08 8.617E−7 6.08 3.035E−7 6.10 1/32 2.267E−8 6.82 6.963E−9 7.02 4.212E−8 4.35 1.088E−8 4.80
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表 7 问题3当
$\tau = 0.001,\;t = 0.1$ 时，对不同$ \alpha $ 的$ {L_\infty } $ 误差Table 7.
$ {L_\infty } $ errors for$\tau = 0.001,\;t = 0.1$ with different$ \alpha $ values in problem 3$ \alpha $ $ h = 0.1 $ $h = 0.062\;5$ DHOC[20] present scheme DHOC[20] present scheme $ {10^{ - 1}} $ 4.764E−4 6.692E−4 7.195E−5 1.070E−4 $ {10^{ - 2}} $ 1.264E−4 1.195E−4 3.955E−5 2.632E−5 $ {10^{ - 3}} $ 1.914E−6 1.875E−6 6.366E−7 4.645E−7 $ {10^{ - 4}} $ 1.996E−8 1.963E−8 6.670E−9 4.937E−9 $ {10^{ - 5}} $ 2.004E−10 1.972E−10 6.701E−11 4.967E−11 $ {10^{ - 6}} $ 2.005E−12 1.973E−12 6.704E−13 4.970E−13
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表 8 问题3当
$h = 0.062\;5,\;\alpha {\text{ = }}0.1,\;t = 2$ 时，对不同网格比$\lambda = {\tau \mathord{/ {\vphantom {\tau {{h^2}}}} } {{h^2}}}$ 的$ {L_\infty } $ 误差和$ {L_2} $ 误差Table 8.
$ {L_\infty } $ errors and$ {L_2} $ errors for$h = 0.062\;5,\;\alpha {\text{ = }}0.1,\;t = 2$ with different mesh ratio$\lambda = {\tau \mathord{/ {\vphantom {\tau {{h^2}}}} } {{h^2}}}$ values in problem 3$ \lambda $ DHOC[20] present scheme $ {L_\infty } $ $ {L_2} $ $ {L_\infty } $ $ {L_2} $ 1 4.782E−10 1.400E−10 7.213E−10 2.469E−10 2 5.190E−10 1.519E−10 7.231E−10 2.477E−10 4 4.547E−7 1.143E−7 7.255E−10 2.490E−10 6 1.212 2.945E−1 7.468E−10 2.567E−10 8 1.093E+1 2.322 7.326E−10 2.529E−10
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