The 4th- and 6th-Order Richardson Extrapolation Methods for Solving 3D Nonlinear Nerve Conduction Equations
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摘要: 该文对一类非线性神经传播方程建立了一类交替方向隐式(ADI)紧致差分方法. 其在时间上有二阶精度, 在空间上有四阶精度. 运用Fourier分析法和能量法可证该方法是无条件线性稳定的. 此外, 对这类方法, 该文提出了两类Richardson外推方法, 以便分别获得时、空方向均有四阶或者六阶精度的外推解, 节省了计算成本. 数值结果验证了该方法的精度和有效性.
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关键词:
- ADI法 /
- 紧致差分法 /
- Richardson外推法 /
- 稳定性
Abstract: An alternating direction implicit (ADI) compact finite difference method (CFDM) was proposed for the numerical solution of the nonlinear nerve conduction equations. The method has 2nd-order accuracy in time and 4th-order accuracy in space, respectively. With the Fourier method and the discrete energy method, the proposed method was proved to be unconditionally linearly stable. In addition, 2 kinds of Richardson extrapolation methods used along with this ADI CFDM, are also developed to get time-space numerical extrapolation solutions with 4th-order or 6th-order accuracy, respectively, and improve the computational efficiency. Numerical results verify the accuracy and efficiency of the proposed method. -
表 1 在t=1时刻,运用ADI紧致法获得的数值结果(Δt=h2)
Table 1. Case 1: numerical results at t=1 with the compact ADI method(Δt=h2)
h Eu R∞ Lu RL2 Ev R∞ Lv RL2 tCPU/s 1/22 1.062 7E-3 - 3.745 6E-4 - 1.033 8E-4 - 3.605 9E-5 - 0.091 9 1/23 6.808 5E-5 3.962 8 2.402 2E-5 3.962 8 6.783 4E-6 3.929 8 2.378 2E-6 3.922 4 0.538 1/24 4.262 0E-6 3.997 7 1.503 8E-6 3.997 6 4.252 6E-7 3.995 6 1.491 5E-7 3.995 0 15.339 1/25 2.664 0E-7 3.999 8 9.400 0E-8 3.999 8 2.658 4E-8 3.999 7 9.323 8E-9 3.999 7 539.05 表 2 例1: 在t=1时刻,运用Richardson-Ⅰ获得的数值结果(Δt=h)
Table 2. Case 1: numerical results at t=1 with the Richardson-Ⅰ method(Δt=h)
h Eu R∞ Lu RL2 Ev R∞ Lv RL2 tCPU/s 1/22 1.115 9E-3 - 3.972 7E-4 - 1.763 4E-4 - 6.240 0E-5 - 0.018 1/23 9.796 7E-5 3.509 7 3.553 8E-5 3.482 7 1.955 4E-5 3.172 8 7.294 2E-6 3.096 7 0.218 1/24 6.691 8E-6 3.871 8 2.460 2E-6 3.852 5 1.394 4E-6 3.809 8 5.377 3E-7 3.761 8 2.836 1/25 4.269 1E-7 3.970 4 1.579 1E-7 3.961 6 8.958 8E-8 3.960 2 3.506 1E-8 3.938 9 47.01 表 3 例1: 在t=1时刻,运用Richardson-Ⅱ获得的数值解(Δt=h)
Table 3. Case 1: numerical results at t=1 with the Richardson-Ⅱ method(Δt=h)
h Eu R∞ Lu RL2 Ev R∞ Lv RL2 tCPU/s 1/22 3.010 6E-5 - 1.139 7E-5 - 9.101 4E-6 - 3.610 3E-6 - 0.238 1/23 6.067 8E-7 5.632 7 2.559 2E-7 5.476 8 1.879 4E-7 5.597 8 8.791 1E-8 5.359 9 2.989 1/24 9.788 9E-9 5.953 9 4.506 4E-9 5.827 6 3.577 3E-9 5.715 2 1.602 9E-9 5.777 3 49.77 1/25 1.672 1E-10 5.871 4 7.503 7E-11 5.908 2 6.524 8E-11 5.776 8 2.683 9E-11 5.900 2 810.65 表 4 例2: 在t=1时刻,运用紧ADI法获得的数值结果(Δt=h2)
Table 4. Case 2: numerical results at t=1 with the compact ADI method(Δt=h2)
h Eu R∞ Lu RL2 Ev R∞ Lv RL2 tCPU/s 1/3 7.564 1E-2 - 3.445 5E-2 - 8.157 9E-2 - 3.737 3E-2 - 0.084 8 1/6 7.487 1E-3 3.336 7 2.632 5E-3 3.710 2 7.806 1E-3 3.385 5 2.748 5E-3 3.765 3 0.143 1/12 4.733 5E-4 3.983 4 1.668 9E-4 3.979 5 4.922 8E-4 3.987 1 1.737 3E-4 3.983 7 2.646 1/24 2.960 5E-5 3.999 0 1.044 0E-5 3.998 7 3.078 4E-5 3.999 2 1.086 5E-5 3.999 0 74.23 表 5 例2: 在t=1时刻,运用Richardson-Ⅰ获得的数值结果(Δt=h)
Table 5. Case 2: numerical results at t=1 with the Richardson-Ⅰ method(Δt=h)
h Eu R∞ Lu RL2 Ev R∞ Lv RL2 tCPU/s 1/3 9.214 4E-2 - 4.175 2E-2 - 9.342 8E-2 - 4.240 8E-2 - 0.037 1/6 1.946 7E-2 2.242 9 6.799 2E-3 2.618 4 1.727 5E-2 2.435 2 6.023 3E-3 2.815 7 0.085 1/12 1.750 2E-3 3.475 4 6.003 7E-4 3.501 4 1.459 8E-3 3.564 9 4.993 5E-4 3.592 4 0.766 1/24 1.213 8E-4 3.849 9 4.142 2E-5 3.857 4 9.910 5E-5 3.880 7 3.373 8E-5 3.887 6 9.97 1/48 7.792 5E-6 3.961 3 2.656 2E-6 3.962 9 6.333 9E-6 3.967 8 2.148 9E-6 3.972 7 164.89 表 6 例2: 在t=1时刻,运用Richardson-Ⅱ获得的数值解(Δt=h)
Table 6. Case 2: numerical results at t=1 with the Richardson-Ⅱ method(Δt=h)
h Eu R∞ Lu RL2 Ev R∞ Lv RL2 tCPU/s 1/3 9.434 3E-3 - 4.147 1E-3 - 7.605 2E-3 - 3.316 6E-3 - 0.085 1/6 5.453 2E-4 4.112 7 1.845 6E-4 4.489 9 3.928 9E-4 4.274 8 1.293 9E-4 4.679 9 0.825 1/12 1.279 1E-5 5.413 9 4.174 9E-6 5.466 2 8.392 8E-6 5.548 8 2.714 0E-6 5.575 2 10.44 1/24 2.251 1E-7 5.828 4 7.248 6E-8 5.847 9 1.725 3E-7 5.604 2 4.957 5E-8 5.774 6 186.29 -
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