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一种求解三维非稳态对流扩散反应方程的高精度有限差分格式

魏剑英 葛永斌

魏剑英,葛永斌. 一种求解三维非稳态对流扩散反应方程的高精度有限差分格式 [J]. 应用数学和力学,2022,43(2):187-197 doi: 10.21656/1000-0887.420151
引用本文: 魏剑英,葛永斌. 一种求解三维非稳态对流扩散反应方程的高精度有限差分格式 [J]. 应用数学和力学,2022,43(2):187-197 doi: 10.21656/1000-0887.420151
WEI Jianying, GE Yongbin. A High-Order Finite Difference Scheme for 3D Unsteady Convection Diffusion Reaction Equations[J]. Applied Mathematics and Mechanics, 2022, 43(2): 187-197. doi: 10.21656/1000-0887.420151
Citation: WEI Jianying, GE Yongbin. A High-Order Finite Difference Scheme for 3D Unsteady Convection Diffusion Reaction Equations[J]. Applied Mathematics and Mechanics, 2022, 43(2): 187-197. doi: 10.21656/1000-0887.420151

一种求解三维非稳态对流扩散反应方程的高精度有限差分格式

doi: 10.21656/1000-0887.420151
基金项目: 国家自然科学基金 (12161067;11961054;11902170);宁夏自然科学基金 (2020AAC03059);宁夏自治区青年拔尖人才培养工程项目
详细信息
    作者简介:

    魏剑英(1980—),女,副教授,硕士(E-mail:weiwy2002@163.com

    葛永斌(1975—),男,教授,博士,博士生导师(通讯作者. E-mail:gyb@nxu.edu.cn

  • 中图分类号: O241.82

A High-Order Finite Difference Scheme for 3D Unsteady Convection Diffusion Reaction Equations

  • 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。

  • 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 $$ h $C-N schemeBTCS schemepresent scheme
    $ {L_\infty } $$\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
    下载: 导出CSV

    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
    下载: 导出CSV

    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}}}$$ {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 $$ \lambda $number of steps in the time directionC-N schemeBTCS schemepresent scheme
    $ {L_\infty } $$ {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
    下载: 导出CSV

    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 2

    schemeLL2
    $ 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
    下载: 导出CSV

    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
    下载: 导出CSV

    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 $
    1/89.079E−53.060E−55.848E−52.078E−5
    1/162.555E−65.159.067E−75.088.617E−76.083.035E−76.10
    1/322.267E−86.826.963E−97.024.212E−84.351.088E−84.80

    下载: 导出CSV

    表  7  问题3当$\tau = 0.001,\;t = 0.1$时,对不同$ \alpha $$ {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 schemeDHOC[20]present scheme
    $ {10^{ - 1}} $4.764E−46.692E−47.195E−51.070E−4
    $ {10^{ - 2}} $1.264E−41.195E−43.955E−52.632E−5
    $ {10^{ - 3}} $1.914E−61.875E−66.366E−74.645E−7
    $ {10^{ - 4}} $1.996E−81.963E−86.670E−94.937E−9
    $ {10^{ - 5}} $2.004E−101.972E−106.701E−114.967E−11
    $ {10^{ - 6}} $2.005E−121.973E−126.704E−134.970E−13
    下载: 导出CSV

    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}}}$$ {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} $
    14.782E−101.400E−107.213E−102.469E−10
    25.190E−101.519E−107.231E−102.477E−10
    44.547E−71.143E−77.255E−102.490E−10
    61.2122.945E−17.468E−102.567E−10
    81.093E+12.3227.326E−102.529E−10
    下载: 导出CSV
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出版历程
  • 收稿日期:  2021-06-03
  • 修回日期:  2021-08-13
  • 网络出版日期:  2022-01-08
  • 刊出日期:  2022-02-01

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