#### 期刊菜单

Riemann-Liouville型分数阶扩散方程的显–隐和隐–显差分方法及数值分析
Explicit-Implicit and Implicit-Explicit Difference Methods and Numerical Analysis for Riemann-Liouville Type Fractional Diffusion Equation

Abstract: For the Riemann-Liouville (R-L) type fractional differential equations describing sub-diffusion phenomena, a kind of explicit-implicit and implicit-explicit difference schemes for numerically solving this problem is constructed. It takes advantage of the fast computation of explicit formats and the unconditional stability of implicit formats, alternatingly applies the classical explicit format and the implicit format by time layer. Then, using the Fourier method analysis, the format is unconditionally stable and convergent. The numerical test results are consistent with the theoret-ical analysis results, and show that the computational efficiency of the explicit-implicit and implic-it-explicit formats is better than the classical implicit format. The explicit-implicit and implic-it-explicit methods are feasible to solve the R-L fractional diffusion equation.

1. 引言

$\frac{\partial u\left(x,t\right)}{\partial t}={}_{0}D{}_{t}^{1-\gamma }\left[\frac{{\partial }^{2}u\left(x,t\right)}{\partial {x}^{2}}\right]+f\left(x,t\right),\text{}t\ge 0$ . (1)

${}_{0}D{}_{t}^{1-\gamma }u\left(x,t\right)=\frac{1}{\Gamma \left(\gamma \right)}\frac{\partial }{\partial t}{\int }_{0}^{t}\frac{u\left(x,\tau \right)}{{\left(t-\tau \right)}^{1-\gamma }}\text{d}\tau$ .

2. 分数阶扩散方程显–隐和隐–显方法

2.1. R-L型时间分数阶扩散方程

$\frac{\partial u\left(x,t\right)}{\partial t}={}_{0}D{}_{t}^{1-\gamma }\left[\frac{{\partial }^{2}u\left(x,t\right)}{\partial {x}^{2}}\right]+f\left(x,t\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}0 . (2)

${}_{0}D{}_{t}^{\gamma -1}\left(\frac{\partial u\left(x,t\right)}{\partial t}\right)={}_{0}D{}_{t}^{\gamma -1}\left\{{}_{0}D{}_{t}^{1-\gamma }\left[{K}_{\gamma }\frac{{\partial }^{2}u\left(x,t\right)}{\partial {x}^{2}}\right]+f\left(x,t\right)\right\}$ ,(3)

${}_{0}{}^{C}D{}_{t}^{\gamma }\frac{\partial u\left(x,t\right)}{\partial t}={K}_{\gamma }\frac{{\partial }^{2}u\left(x,t\right)}{\partial {x}^{2}}-{\left\{{}_{0}D{}_{t}^{-\gamma }\left[{K}_{\gamma }\frac{{\partial }^{2}u\left(x,t\right)}{\partial {x}^{2}}\right]\right\}|}_{t=0}+{}_{0}D{}_{t}^{\gamma -1}f\left(x,t\right)$ , (4)

${\left\{{}_{0}D{}_{t}^{-\gamma }\left[{K}_{\gamma }\frac{{\partial }^{2}u\left(x,t\right)}{\partial {x}^{2}}\right]\right\}|}_{t=0}=0$ .

${}_{0}^{C}{D}_{t}^{\gamma }\frac{\partial u\left(x,t\right)}{\partial t}={K}_{\gamma }\frac{{\partial }^{2}u\left(x,t\right)}{\partial {x}^{2}}+g\left(x,t\right),g\left(x,t\right)={}_{0}D{}_{t}^{\gamma -1}f\left(x,t\right)$ (5)

2.2. 显–隐和隐–显格式的构造

${t}_{k}=k\tau ,k=0,1,\cdots ,N$${x}_{j}=jh,j=0,1,\cdots ,M$$\tau =T/N$$h=L/M$。定义网格函数

${u}_{i}^{n}=u\left({x}_{i},{t}_{n}\right)$ , ${f}_{i}^{n}=f\left({x}_{i},{t}_{n}\right)$ , ${g}_{i}^{n}=g\left({x}_{i},{t}_{n}\right)$ , $\left({x}_{i},{t}_{n}\right)\in \left[0,L\right]×\left[0,T\right]$ .

${D}_{\tau }^{\gamma }{w}^{n}=\frac{{\tau }^{-\gamma }}{\Gamma \left(2-\gamma \right)}\left[{w}^{n}-\underset{k=1}{\overset{n-1}{\sum }}\left({a}_{n-k-1}-{a}_{n-k}\right){w}^{k}-{a}_{n-1}{w}^{0}\right]$ , ${a}_{k}={\left(k+1\right)}^{1-\alpha }-{k}^{1-\alpha }$ .

$\begin{array}{l}\frac{1}{\Gamma \left(1-\alpha \right)}{\int }_{0}^{{t}_{n}}\frac{{y}^{\prime }\left(\xi \right)\text{d}\xi }{{\left(t-\xi \right)}^{\alpha }}-\frac{{\tau }^{-\alpha }}{\Gamma \left(2-\alpha \right)}\left[{y}^{n}-\underset{k=1}{\overset{n-1}{\sum }}\left({a}_{n-k-1}-{a}_{n-k}\right){y}^{k}-{a}_{n-1}{y}^{0}\right]\\ \le \frac{1}{\Gamma \left(2-\alpha \right)}\left[\frac{1-\alpha }{12}+\frac{{2}^{2-\alpha }}{2-\alpha }-\left(1+{2}^{\alpha }\right)\right]\underset{0\le t\le {t}_{n}}{\mathrm{max}}|{y}^{″}\left(t\right)|{\tau }^{2-\alpha }\end{array}$ .

$\frac{{\tau }^{-\gamma }}{\Gamma \left(2-\gamma \right)}\left[{u}_{j}^{k}-\underset{i=1}{\overset{k-1}{\sum }}\left({a}_{k-i-1}-{a}_{k-i}\right){u}_{j}^{i}-{a}_{k-1}{u}_{j}^{0}\right]={\delta }_{x}^{2}{u}_{j}^{k}+{g}_{j}^{k}$ ,(6)

$\frac{{\tau }^{-\gamma }}{\Gamma \left(2-\gamma \right)}\left[{u}_{j}^{k}-\underset{i=1}{\overset{k-1}{\sum }}\left({a}_{k-i-1}-{a}_{k-i}\right){u}_{j}^{i}-{a}_{k-1}{u}_{j}^{0}\right]={\delta }_{x}^{2}{u}_{j}^{k-1}+{g}_{j}^{k},\text{\hspace{0.17em}}\text{\hspace{0.17em}}j=1,2,\cdots ,M-1$ . (7)

$\begin{array}{l}{u}_{0}^{k}=\phi \left({t}_{k}\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}{u}_{m}^{k}=\psi \left({t}_{k}\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}k=0,1,\cdots ,N\\ {u}_{j}^{0}=w\left({x}_{j}\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}j=0,1,\cdots ,M\end{array}$

$\left\{\begin{array}{l}\frac{{\tau }^{-\gamma }}{\Gamma \left(2-\gamma \right)}\left[{u}_{j}^{2k+1}-\underset{i=1}{\overset{2k}{\sum }}\left({a}_{2k-i}-{a}_{2k-i+1}\right){u}_{j}^{i}-{a}_{2k}{u}_{j}^{0}\right]={\delta }_{x}^{2}{u}_{j}^{2k}+{g}_{j}^{2k+1},\\ \frac{{\tau }^{-\gamma }}{\Gamma \left(2-\gamma \right)}\left[{u}_{j}^{2k+2}-\underset{i=1}{\overset{2k+1}{\sum }}\left({a}_{2k-i+1}-{a}_{2k-i}\right){u}_{j}^{i}-{a}_{2k+1}{u}_{j}^{0}\right]={\delta }_{x}^{2}{u}_{j}^{2k+2}+{g}_{j}^{2k+2},\end{array}$ (8)

$\left\{\begin{array}{l}\frac{{\tau }^{-\gamma }}{\Gamma \left(2-\gamma \right)}\left[{u}_{j}^{2k+1}-\underset{i=1}{\overset{2k}{\sum }}\left({a}_{2k-i}-{a}_{2k-i+1}\right){u}_{j}^{i}-{a}_{2k}{u}_{j}^{0}\right]={\delta }_{x}^{2}{u}_{j}^{2k+1}+{g}_{j}^{2k+1},\\ \frac{{\tau }^{-\gamma }}{\Gamma \left(2-\gamma \right)}\left[{u}_{j}^{2k+2}-\underset{i=1}{\overset{2k+1}{\sum }}\left({a}_{2k-i+1}-{a}_{2k-i}\right){u}_{j}^{i}-{a}_{2k+1}{u}_{j}^{0}\right]={\delta }_{x}^{2}{u}_{j}^{2k+1}+{g}_{j}^{2k+2}.\end{array}$ (9)

3. 显–隐和隐–显方法的稳定性分析

$\left\{\begin{array}{l}{u}_{j}^{2k+1}=\left({a}_{0}-{a}_{1}-2\mu \right){u}_{j}^{2k}+\mu {u}_{j+1}^{2k}+\mu {u}_{j-1}^{2k}+\underset{i=1}{\overset{2k-1}{\sum }}\left({a}_{2k-i}-{a}_{2k-i+1}\right){u}_{j}^{i}-{a}_{2k}{u}_{j}^{0}+\mu {g}_{j}^{2k+1},\\ \left(1+2\mu \right){u}_{j}^{2k+2}-\mu {u}_{j+1}^{2k+2}-\mu {u}_{j-1}^{2k+2}=\left({a}_{0}-{a}_{1}\right){u}_{j}^{2k+1}+\underset{i=1}{\overset{2k}{\sum }}\left({a}_{2k-i+1}-{a}_{2k-i+2}\right){u}_{j}^{i}-{a}_{2k+1}{u}_{j}^{0}+\mu {g}_{j}^{2k+2},\end{array}$ (10)

${U}_{j}^{k}$ 是显–隐格式的近似解，定义 ${\rho }_{j}^{k}={u}_{j}^{k}-{U}_{j}^{k}$ $\left(k=1,2,\cdots ,N,j=1,2,\cdots ,M-1\right)$${\rho }^{k}={\left[{\rho }_{1}^{k},{\rho }_{2}^{k},\cdots ,{\rho }_{M-1}^{k}\right]}^{\text{T}}$，代入(10)式可得如下误差扰动方程：

$\left\{\begin{array}{l}{\rho }_{j}^{2k+1}=\left({a}_{0}-{a}_{1}-2\mu \right){\rho }_{j}^{2k}+\mu {\rho }_{j+1}^{2k}+\mu {\rho }_{j-1}^{2k}+\underset{i=1}{\overset{2k-1}{\sum }}\left({a}_{2k-i}-{a}_{2k-i+1}\right){\rho }_{j}^{i}-{a}_{2k}{\rho }_{j}^{0},\\ \left(1+2\mu \right){\rho }_{j}^{2k+2}-\mu {\rho }_{j+1}^{2k+2}-\mu {\rho }_{j-1}^{2k+2}=\left({a}_{0}-{a}_{1}\right){\rho }_{j}^{2k+1}+\underset{i=1}{\overset{2k}{\sum }}\left({a}_{2k-i+1}-{a}_{2k-i+2}\right){\rho }_{j}^{i}-{a}_{2k+1}{\rho }_{j}^{0},\end{array}$ (11)

${\rho }^{k}\left(x\right)=\left\{\begin{array}{l}0,\text{}0\le x\le \frac{h}{2}\text{,}\\ {\rho }_{j}^{k},\text{}{x}_{j}-\frac{h}{2} $\left(k=1,2,\cdots ,N\right)$

${\rho }^{k}\left(x\right)$ 的Fourier展开形式为：

${\rho }^{k}\left(x\right)=\underset{l=-\infty }{\overset{\infty }{\sum }}{d}_{k}\left(l\right){\text{e}}^{i2\pi lx/L}$ ,

$\begin{array}{c}{‖{\rho }^{k}‖}_{2}^{2}=\underset{j=1}{\overset{M-1}{\sum }}h{|{\rho }_{j}^{k}|}^{2}\\ ={\int }_{0}^{\frac{h}{2}}{|{\rho }^{k}\left(x\right)|}^{2}\text{d}x+\underset{j=1}{\overset{M-1}{\sum }}{\int }_{{x}_{j}-\frac{h}{2}}^{{x}_{j}+\frac{h}{2}}{|{\rho }^{k}\left(x\right)|}^{2}\text{d}x+{\int }_{L-\frac{h}{2}}^{L}{|{\rho }^{k}\left(x\right)|}^{2}\text{d}x\\ ={\int }_{0}^{L}{|{\rho }^{k}\left(x\right)|}^{2}\text{d}x\end{array}$ (12)

${\int }_{0}^{L}{|{\rho }^{k}\left(x\right)|}^{2}\text{d}x=L\underset{m=-\infty }{\overset{\infty }{\sum }}{|{d}_{k}\left(m\right)|}^{2}$，即 ${‖{\rho }^{k}‖}_{2}^{2}=L\underset{m=-\infty }{\overset{\infty }{\sum }}{|{d}_{k}\left(m\right)|}^{2}$ .

$m=1$ 时(m是时间层数)，我们使用显式差分格式，有

${\rho }_{j}^{1}=\left(1-2\mu \right){\rho }_{j}^{0}+\mu {\rho }_{j+1}^{0}+\mu {\rho }_{j-1}^{0}$ ,

${d}_{1}=\left(\mu {\text{e}}^{i\sigma h}+1-2\mu +\mu {\text{e}}^{-i\sigma h}\right){d}_{0}$ ,

${\text{e}}^{±i\theta }=\mathrm{cos}\theta ±i\mathrm{sin}\theta$，可得

$|{d}_{1}|=|1-4\mu {\mathrm{sin}}^{2}\frac{\sigma h}{2}||{d}_{0}|\le C|{d}_{0}|$ , (13)

$m=2$ 时，第一层使用显式格式，第二层使用隐式格式，

${\rho }_{j}^{1}=\left(1-2\mu \right){\rho }_{j}^{0}+\mu {\rho }_{j+1}^{0}+\mu {\rho }_{j-1}^{0}$ ,

$\left(1+2\mu \right){\rho }_{j}^{2}-\mu {\rho }_{j+1}^{2}-\mu {\rho }_{j-1}^{2}=\left({a}_{0}-{a}_{1}\right){\rho }_{j}^{1}+{a}_{1}{\rho }_{j}^{0}$ ,

$\left(1+4\mu {\mathrm{sin}}^{2}\frac{\sigma h}{2}\right){d}_{2}=\left({a}_{0}-{a}_{1}\right)\left(1-4\mu {\mathrm{sin}}^{2}\frac{\sigma h}{2}\right){d}_{0}+{a}_{1}{d}_{0}$ ,

$|{d}_{2}|=|\frac{\left(1-{a}_{1}\right)\left(1-4\mu {\mathrm{sin}}^{2}\frac{\sigma h}{2}\right)+{a}_{1}}{1+4\mu {\mathrm{sin}}^{2}\frac{\sigma h}{2}}||{d}_{0}|\le |\frac{1-\left(1-{a}_{1}\right)4\mu {\mathrm{sin}}^{2}\frac{\sigma h}{2}}{1+4\mu {\mathrm{sin}}^{2}\frac{\sigma h}{2}}||{d}_{0}|\le C|{d}_{0}|$ .(14)

$m=2k$ 时，假设 $|{d}_{n}|\le C|{d}_{0}|,n\le 2k$ 成立；当 $m=2k+2$，有

$\left\{\begin{array}{l}{d}_{2k+1}=\left({a}_{0}-{a}_{1}-4\mu {\mathrm{sin}}^{2}\frac{\sigma h}{2}\right){d}_{2k}+\underset{i=1}{\overset{2k-1}{\sum }}\left({a}_{2k-i}-{a}_{2k-i+1}\right){d}_{i}-{a}_{2k}{d}_{0},\\ \left(1+4\mu {\mathrm{sin}}^{2}\frac{\sigma h}{2}\right){d}_{2k+2}=\left({a}_{0}-{a}_{1}\right){d}_{2k+1}+\underset{i=1}{\overset{2k}{\sum }}\left({a}_{2k-i+1}-{a}_{2k-i+2}\right){d}_{j}-{a}_{2k+1}{d}_{0},\end{array}$

$\begin{array}{l}\left(1+4\mu {\mathrm{sin}}^{2}\frac{\sigma h}{2}\right){d}_{2k+2}\\ =\left({a}_{0}-{a}_{1}\right)\left[\left({a}_{0}-{a}_{1}-4\mu {\mathrm{sin}}^{2}\frac{\sigma h}{2}\right){d}_{2k}+\underset{i=1}{\overset{2k-1}{\sum }}\left({a}_{2k-i}-{a}_{2k-i+1}\right){d}_{i}-{a}_{2k}{d}_{0}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{ }+\underset{i=1}{\overset{2k}{\sum }}\left({a}_{2k-i+1}-{a}_{2k-i+2}\right){d}_{j}-{a}_{2k+1}{d}_{0}\end{array}$

$\stackrel{˜}{\mu }=4\mu {\mathrm{sin}}^{2}\frac{\sigma h}{2}$，由上式可得

$\begin{array}{l}\left(1+4\mu {\mathrm{sin}}^{2}\frac{\sigma h}{2}\right)|{d}_{2k+2}|\\ \le \left\{\left({a}_{0}-{a}_{1}\right)\left[|\left({a}_{0}-{a}_{1}\right)+4\mu {\mathrm{sin}}^{2}\frac{\sigma h}{2}|+\underset{i=1}{\overset{2k-1}{\sum }}|{a}_{2k-i}-{a}_{2k-i+1}|+|{a}_{2k}|\right]+\underset{i=1}{\overset{2k}{\sum }}|{a}_{2k-i+1}-{a}_{2k-i+2}|+|{a}_{2k+1}|\right\}C|{d}_{0}|\\ \le \left\{\left(1-{a}_{1}\right)\left(1+4\mu {\mathrm{sin}}^{2}\frac{\sigma h}{2}\right)+{a}_{1}\right\}C|{d}_{0}|\\ \le \left[1+\left(1-{a}_{1}\right)4\mu {\mathrm{sin}}^{2}\frac{\sigma h}{2}\right]C|{d}_{0}|\end{array}$

$|{d}_{2k+2}|\le \frac{1+4\mu {\mathrm{sin}}^{2}\frac{\sigma h}{2}}{1+\left(1-{a}_{1}\right)4\mu {\mathrm{sin}}^{2}\frac{\sigma h}{2}}C|{d}_{0}|\le C|{d}_{0}|$ . (15)

${‖{\rho }^{k}‖}^{2}=L\underset{m=-\infty }{\overset{\infty }{\sum }}{|{d}_{k}\left(m\right)|}^{2}\le CL\underset{m=-\infty }{\overset{\infty }{\sum }}{|{d}_{0}\left(m\right)|}^{2}=C{‖{\rho }^{0}‖}^{2},\text{\hspace{0.17em}}\text{\hspace{0.17em}}k=1,2,\cdots ,N$ .

4. 显–隐和隐–显方法的收敛性分析

$\begin{array}{l}\frac{1}{\Gamma \left(1-\alpha \right)}{\int }_{0}^{{t}_{n}}\frac{{y}^{\prime }\left(\xi \right)\text{d}\xi }{{\left(t-\xi \right)}^{\alpha }}-\frac{{\tau }^{-\alpha }}{\Gamma \left(2-\alpha \right)}\left[{y}^{n}-\underset{k=1}{\overset{n-1}{\sum }}\left({a}_{n-k-1}-{a}_{n-k}\right){y}^{k}-{a}_{n-1}{y}^{0}\right]\\ \le \frac{1}{\Gamma \left(2-\alpha \right)}\left[\frac{1-\alpha }{12}+\frac{{2}^{2-\alpha }}{2-\alpha }-\left(1+{2}^{\alpha }\right)\right]\underset{0\le t\le {t}_{n}}{\mathrm{max}}|{y}^{″}\left(t\right)|{\tau }^{2-\alpha }\end{array}$ (16)

$\frac{{\tau }^{-\gamma }}{\Gamma \left(2-\gamma \right)}\left[{u}_{j}^{2k+1}-\underset{i=1}{\overset{2k}{\sum }}\left({a}_{2k-i}-{a}_{2k-i+1}\right){u}_{j}^{i}-{a}_{2k}{u}_{j}^{0}\right]=\left(1-\theta \right){\delta }_{x}^{2}{u}_{j}^{2k}+\theta {\delta }_{x}^{2}{u}_{j}^{2k}+{g}_{j}^{2k+1}$ , (17)

${R}_{j}^{2k+1}=\frac{{\tau }^{-\gamma }}{\Gamma \left(2-\gamma \right)}\left[{u}_{j}^{2k+1}-\underset{i=1}{\overset{2k}{\sum }}\left({a}_{2k-i}-{a}_{2k-i+1}\right){u}_{j}^{i}-{a}_{2k}{u}_{j}^{0}\right]-\left(1-\theta \right){\delta }_{x}^{2}{u}_{j}^{2k}-\theta {\delta }_{x}^{2}{u}_{j}^{2k}-{g}_{j}^{2k+1}$ (18)

$\theta {K}_{\alpha }{\delta }_{x}^{2}{u}_{j}^{k}+\left(1-\theta \right){K}_{\alpha }{\delta }_{x}^{2}{u}_{j}^{k-1}={u}_{xx}+\frac{{h}_{}^{2}}{12}{u}_{xxxx}+\left(1-\theta \right)\tau {u}_{xxt}+O\left({\tau }^{2-\alpha }+{h}^{2}\right)$ (19)

$\begin{array}{c}{R}_{j}^{2k+1}=\frac{{\tau }^{-\gamma }}{\Gamma \left(2-\gamma \right)}\left[{u}_{j}^{2k+1}-\underset{i=1}{\overset{2k}{\sum }}\left({a}_{2k-i}-{a}_{2k-i+1}\right){u}_{j}^{i}-{a}_{2k}{u}_{j}^{0}\right]-\left(1-\theta \right){\delta }_{x}^{2}{u}_{j}^{2k}-\theta {\delta }_{x}^{2}{u}_{j}^{2k}-{g}_{j}^{2k+1}\\ ={}_{0}{}^{C}D{}_{\tau }^{\alpha }u\left({x}_{i},{t}_{k+1}\right)-\left(1-\theta \right)\tau {}_{0}{}^{C}D{}_{\tau }^{\alpha +1}u\left({x}_{i},{t}_{k+1}\right)-{u}_{xx}+\frac{{h}^{2}}{12}{u}_{xxxx}+\left(1-\theta \right)\tau {u}_{xxt}+O\left({\tau }^{2-\alpha }+{h}^{2}\right)\\ =O\left({\tau }^{2-\alpha }+{h}^{2}\right)\end{array}$

$|{R}_{j}^{k}|\le {c}_{1}\left({\tau }^{2-\gamma }+{h}^{2}\right)$ , $k=1,2,\cdots ,N,j=1,2,\cdots ,M-1$ . (20)

${e}_{j}^{k}=u\left({x}_{j},{t}_{k}\right)-{u}_{j}^{k}$ $\left(k=1,2,\cdots ,N,j=1,2,\cdots ,M-1\right)$ .

${e}^{k}={\left[{e}_{1}^{k},{e}_{2}^{k},\cdots ,{e}_{M-1}^{k}\right]}^{\text{T}}$ , ${R}^{k}={\left[{R}_{1}^{k},{R}_{2}^{k},\cdots ,{R}_{M-1}^{k}\right]}^{\text{T}}$ .

${e}^{k}\left(x\right)=\left\{\begin{array}{l}0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\le x\le \frac{h}{2},\\ {e}_{j}^{k},\text{\hspace{0.17em}}\text{\hspace{0.17em}}{x}_{j}-\frac{h}{2} $\left(k=1,2,\cdots ,N\right)$

${R}^{k}\left(x\right)=\left\{\begin{array}{l}0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\le x\le \frac{h}{2},\\ {R}_{j}^{k},\text{\hspace{0.17em}}\text{\hspace{0.17em}}{x}_{j}-\frac{h}{2} $\left(k=1,2,\cdots ,N\right)$

${e}^{k}\left(x\right),{R}^{k}\left(x\right)$ 的Fourier展开形式为：

${e}^{k}\left(x\right)=\underset{l=-\infty }{\overset{\infty }{\sum }}{\eta }_{k}\left(l\right){\text{e}}^{i2\pi lx/L}$，其中系数 ${\eta }_{k}\left(l\right)=\frac{1}{L}{\int }_{0}^{L}{e}^{k}\left(x\right){\text{e}}^{-i2\pi lx/L}\text{d}x$

${R}^{k}\left(x\right)=\underset{l=-\infty }{\overset{\infty }{\sum }}{\xi }_{k}\left(l\right){\text{e}}^{i2\pi lx/L}$，其中系数 ${\xi }_{k}\left(l\right)=\frac{1}{L}{\int }_{0}^{L}{R}^{k}\left(x\right){\text{e}}^{-i2\pi lx/L}\text{d}x$

,(21)

. (22)

,

,

(23)

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(24)

(22)式等式右边的系数数列是收敛的，因此存在一个正数使得

. (25)

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5. 数值试验

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Figure 1. Surfaces of analytic solution and E-I, I-E scheme numerical solutions

Table 1. Accuracy comparison analysis of three numerical schemes

Table 2. Time direction convergence order of three difference schemes (h = 64)

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Table 3. Spatial direction convergence order of three difference schemes under different

Figure 2. Surface of numerical solution of three different schemes for different

Table 4. Comparison of calculation times for three differential schemes (unit: s)

Figure 3. Comparison of calculation times for three differential schemes

6. 结论

NOTES

*通讯作者。

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