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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):1-11 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):1-11 doi: 10.21656/1000-0887.420151
WEI Jianying, GE Yongbin. A High-order Finite Difference Scheme for the 3D Unsteady Convection Diffusion Reaction Equation[J]. Applied Mathematics and Mechanics. doi: 10.21656/1000-0887.420151
Citation: WEI Jianying, GE Yongbin. A High-order Finite Difference Scheme for the 3D Unsteady Convection Diffusion Reaction Equation[J]. Applied Mathematics and Mechanics. 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 (11772165;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 the 3D Unsteady Convection Diffusion Reaction Equation

  • 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。最后给出了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 solution on $ z = 1.5 $ in the region $ 1.2 \leqslant x,y \leqslant 1.8 $ for problem 2 when$ h = 0.025{\text{,}}t = 1.25,p = q = r = 0.8{\text{,}}\tau = 6.25 \times {10^{ - 3}} $: (a) PHOC-ADI scheme and exact; (b) EHOC-ADI scheme and exact; (c) RHOC-ADI scheme and exact; (d) the present scheme and exact

    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 $内截面z=1.5${\{z} = 1.5end{document}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 solution on ${{z}} = 1.5$ in the region$ 1.2 \leqslant x,y \leqslant 1.8 $ for problem 2 when $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; (b) EHOC-ADI scheme and exact; (c) RHOC-ADI scheme and exact;(d) the present scheme and exact.

    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 } $ error, $ {L_2} $ error and convergence rate when $\tau = {h^2},t = 0.25$ for 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 ①1/81.60E−25.62E−31.63E−25.73E−32.74E−41.67E−4
    1/163.97E−32.011.40E−32.014.05E−32.001.42E−32.007.81E−65.133.87E−65.43
    1/329.90E−42.003.48E−42.001.01E−32.003.56E−42.003.94E−74.301.62E−74.57
    1/642.47E−42.008.71E−52.002.53E−42.008.89E−52.002.29E−84.108.08E−94.32
    case ②1/81.58E−25.55E−31.62E−25.66E−32.83E−41.70E−4
    1/163.94E−32.001.38E−32.014.02E−32.011.41E−32.018.48E−65.064.10E−65.37
    1/329.83E−42.003.44E−42.001.00E−32.013.52E−42.004.30E−74.301.76E−74.54
    1/642.46E−42.008.61E−52.002.51E−41.998.79E−52.002.52E−84.099.75E−94.17
    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 } $ error, $ {L_2} $ error, convergence rate and CPU time when $h = 0.031\;25,t = 0.5$ for problem 1

    $ p,q,r $$ \tau $$ {L_\infty } $$\delta $$ {L_2} $$\delta $CPU time tCPU/s
    case ①0.14.883E−41.654E−4195.934
    0.051.234E−41.984.076E−52.01275.417
    0.0253.155E−51.971.050E−51.96334.355
    0.01258.050E−61.972.695E−61.96406.406
    0.006252.296E−61.817.981E−71.76455.322
    case ②0.18.220E−42.528E−4240.216
    0.052.017E−42.036.189E−52.03345.248
    0.0255.056E−52.001.550E−52.00426.347
    0.01251.292E−51.973.920E−61.98520.264
    0.006253.519E−61.881.039E−61.92539.716
    下载: 导出CSV

    表  3  问题1当$ h = {1 \mathord{\left/ {\vphantom {1 {32}}} \right. } {32}}{\text{,}}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 } $ error and $ {L_2} $ error when $ h = {1 \mathord{\left/ {\vphantom {1 {32}}} \right. } {32}}{\text{,}}t = 1 $ for different mesh ratio $\lambda = {\tau \mathord{/ {\vphantom {\tau {{h^2}}}} } {{h^2}}}$ for problem 1

    $ p,q,r $$ \lambda $steps in the time directionC-N schemeBTCS schemepresent scheme
    $ {L_\infty } $$ {L_2} $$ {L_\infty } $$ {L_2} $$ {L_\infty } $$ {L_2} $
    case ①0.812802.044E−37.086E−42.079E−37.213E−48.703E−73.611E−7
    1.66402.043E−37.086E−42.114E−37.340E−49.715E−74.008E−7
    3.23202.043E−37.085E−42.185E−37.595E−41.495E−65.737E−7
    6.41602.043E−37.083E−42.327E−38.104E−43.841E−61.330E−6
    case ②0.812802.110E−37.362E−42.144E−37.485E−47.827E−73.002E−7
    1.66402.110E−37.362E−42.178E−37.608E−47.906E−72.496E−7
    3.23202.110E−37.364E−42.246E−37.854E−41.381E−64.394E−7
    6.41602.112E−37.371E−42.383E−38.345E−46.839E−62.142E−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 } $ error and $ {L_2} $ error when $ h = 0.025 $ for 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 } $ error, $ {L_2} $ error and convergence rate when $ t = 0.25,\tau = {h^2},p = q = r = 0.8 $ for problem 2

    $ h $$ {L_\infty } $$\delta $$ {L_2} $$\delta $
    1/402.287E−42.861E−5
    1/604.586E−53.965.516E−64.06
    1/801.470E−53.951.730E−64.03
    1/1006.013E−64.017.057E−74.02
    1/1202.886E−64.033.396E−74.01
    下载: 导出CSV

    表  6  问题3当$ \tau = {h^2}{\text{,}}\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 } $ error, $ {L_2} $ error and convergence rate when $ \tau = {h^2}{\text{,}}\alpha = 0.1,t = 0.5 $ for 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 } $ error when $ \tau = 0.001,t = 0.1 $ for different $ \alpha $ for 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 } $ error and $ {L_2} $ error when $ h = 0.0625,\alpha {\text{ = }}0.1,t = 2 $ for different mesh ratio $\lambda = {\tau \mathord{/ {\vphantom {\tau {{h^2}}}} } {{h^2}}}$ for 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
    61.212E+02.945E−17.468E−102.567E−10
    81.093E+12.322E+07.326E−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

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