﻿ 时间分数阶Fokker-Planck方程有限体积法 A Finite Volume Method for Time Fractional Fokker-Planck Equations

Vol. 08  No. 02 ( 2019 ), Article ID: 28840 , 7 pages
10.12677/AAM.2019.82023

A Finite Volume Method for Time Fractional Fokker-Planck Equations

Lan Huang

School of Mathematics and Statistics, Changsha University of Science and Technology, Changsha Hunan

Received: Jan. 22nd, 2019; accepted: Feb. 6th, 2019; published: Feb. 13th, 2019

ABSTRACT

We present a finite volume method for solving the time fractional Fokker-Planck equations with space-and-time-dependent forcing. Numerical test shows that the convergence rates for time and for space are order 1 and order 2 respectively.

Keywords:Time Fractional Fokker-Planck Equations, Finite Volume Method

1. 研究的问题

$\frac{\partial u}{\partial t}=\left({k}_{\alpha }\frac{{\partial }^{2}}{\partial {x}^{2}}-\frac{\partial }{\partial x}f\right){D}_{t}^{1-\alpha }u,\text{\hspace{0.17em}}a (1.1)

$u\left(x,0\right)=\phi \left(x\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}a (1.2)

2. 离散

$\left[0,T\right]$ 上取分点为 ${t}_{k}=k\Delta t$$k=0,1,\cdots ,L$ ，其中时间步长 $\Delta t=T/L$ ，L为正整数。在区间 $\left[a,b\right]$ 上取分点 ${x}_{i}=a+ih,\text{\hspace{0.17em}}i=0,1,\cdots ,N+1$ ，其中空间步长 $h=\left(b-a\right)/\left(N+1\right)$ ，N为正整数。N个空间有限体为 $\left[{x}_{i-\frac{1}{2}},{x}_{i+\frac{1}{2}}\right],\text{\hspace{0.17em}}i=1,\cdots ,N$ ，其中 ${x}_{i+\frac{1}{2}}=\frac{{x}_{i}+{x}_{i+1}}{2},\text{\hspace{0.17em}}i=0,1,\cdots ,N$

${\frac{\partial u}{\partial t}|}_{{t}_{n}}=\left[{k}_{\alpha }\frac{{\partial }^{2}}{\partial {x}^{2}}-\frac{\partial }{\partial x}f\left(x,{t}_{n}\right)\right]{D}_{t}^{1-\alpha }u\left(x,{t}_{n}\right).$ (2.1)

${\frac{\partial u}{\partial t}|}_{{t}_{n}}=\frac{u\left(x,{t}_{n}\right)-u\left(x,{t}_{n-1}\right)}{\Delta t}+{r}_{n}^{\left(1\right)},$ (2.2)

${D}_{t}^{1-\alpha }u\left(x,{t}_{n}\right)=\Delta {t}^{\alpha -1}{\sum }_{j=0}^{n}{\omega }_{n-j}^{1-\alpha }u\left(x,{t}_{j}\right)+{r}_{n}^{\left(2\right)},$ (2.3)

$\frac{u\left(x,{t}_{n}\right)-u\left(x,{t}_{n-1}\right)}{\Delta t}+{r}_{n}^{\left(1\right)}=\left[{k}_{\alpha }\frac{{\partial }^{2}}{\partial {x}^{2}}-\frac{\partial }{\partial x}f\left(x,{t}_{n}\right)\right]\left[\Delta {t}^{\alpha -1}{\sum }_{j=0}^{n}{\omega }_{n-j}^{1-\alpha }u\left(x,{t}_{j}\right)+{r}_{n}^{\left(2\right)}\right].$ (2.4)

${u}_{n}\left(x\right)$ 表示 $u\left(x,{t}_{n}\right)$ 的近似，由(2.4)我们得到原问题的时间半离散格式

$\frac{{u}_{n}\left(x\right)-{u}_{n-1}\left(x\right)}{\Delta t}=\left[{k}_{\alpha }\frac{{\partial }^{2}}{\partial {x}^{2}}-\frac{\partial }{\partial x}f\left(x,{t}_{n}\right)\right]\left[\Delta {t}^{\alpha -1}{\sum }_{j=0}^{n}{\omega }_{n-j}^{1-\alpha }{u}_{j}\left(x\right)\right],$ (2.5)

${v}_{n}\left(x\right):=\Delta {t}^{\alpha -1}{\sum }_{j=0}^{n}{\omega }_{n-j}^{1-\alpha }{u}_{j}\left(x\right).$ (2.6)

${u}_{n}\left(x\right):=\Delta {t}^{1-\alpha }{\sum }_{j=0}^{n}{\omega }_{n-j}^{\alpha -1}{v}_{j}\left(x\right).$ (2.7)

$\Delta {t}^{-\alpha }\left[{\sum }_{j=0}^{n-1}\left({\omega }_{n-j}^{\alpha -1}-{\omega }_{n-j-1}^{\alpha -1}\right){v}_{j}\left(x\right)+{\omega }_{0}^{\alpha -1}{v}_{n}\left(x\right)\right]=\left[{k}_{\alpha }\frac{{\partial }^{2}}{\partial {x}^{2}}-\frac{\partial }{\partial x}f\left(x,{t}_{n}\right)\right]{v}_{n}\left(x\right).$ (2.8)

${\int }_{{x}_{i-\frac{1}{2}}}^{{x}_{i+\frac{1}{2}}}\Delta {t}^{-\alpha }\left[{\sum }_{j=0}^{n-1}\left({\omega }_{n-j}^{\alpha -1}-{\omega }_{n-j-1}^{\alpha -1}\right){v}_{j}\left(x\right)+{\omega }_{0}^{\alpha -1}{v}_{n}\left(x\right)\right]\text{d}x={\int }_{{x}_{i-\frac{1}{2}}}^{{x}_{i+\frac{1}{2}}}\left[{k}_{\alpha }\frac{{\partial }^{2}}{\partial {x}^{2}}-\frac{\partial }{\partial x}f\left(x,{t}_{n}\right)\right]{v}_{n}\left(x\right)\text{d}x.$ (2.9)

$\begin{array}{l}{\int }_{{x}_{i-\frac{1}{2}}}^{{x}_{i+\frac{1}{2}}}\Delta {t}^{-\alpha }\left[{\sum }_{j=0}^{n-1}\left({\omega }_{n-j}^{\alpha -1}-{\omega }_{n-j-1}^{\alpha -1}\right){v}_{j}\left(x\right)+{\omega }_{0}^{\alpha -1}{v}_{n}\left(x\right)\right]\text{d}x\\ =h\Delta {t}^{-\alpha }\left[{\sum }_{j=0}^{n-1}\left({\omega }_{n-j}^{\alpha -1}-{\omega }_{n-j-1}^{\alpha -1}\right){v}_{i}^{j}+{\omega }_{0}^{\alpha -1}{v}_{i}^{n}\right]-h{r}_{i,n}^{\left(1\right)},\end{array}$ (2.10)

$h|{r}_{i,n}^{\left(1\right)}|\le C{h}^{3}.$ (2.11)

${f}_{i+1/2,n}$ 表示 $\text{}f\left({x}_{i+1/2},{t}_{n}\right)$ ，(2.9)式右侧可以写成

${\int }_{{x}_{i-\frac{1}{2}}}^{{x}_{i+\frac{1}{2}}}\left[{k}_{\alpha }\frac{{\partial }^{2}}{\partial {x}^{2}}-\frac{\partial }{\partial x}f\left(x,{t}_{n}\right)\right]{v}_{n}\left(x\right)\text{d}x={k}_{\alpha }\left({\frac{\partial {v}_{n}}{\partial x}|}_{{x}_{i+\frac{1}{2}}}-{\frac{\partial {v}_{n}}{\partial x}|}_{{x}_{i-\frac{1}{2}}}\right)-\left({f}_{i+\frac{1}{2},n}{v}_{i+\frac{1}{2}}^{n}-{f}_{i-\frac{1}{2},n}{v}_{i-\frac{1}{2}}^{n}\right).$ (2.12)

${k}_{\alpha }\left({\frac{\partial {v}_{n}}{\partial x}|}_{{x}_{i+\frac{1}{2}}}-{\frac{\partial {v}_{n}}{\partial x}|}_{{x}_{i-\frac{1}{2}}}\right)={k}_{\alpha }\left(\frac{{v}_{i+1}^{n}-{v}_{i}^{n}}{h}-\frac{{v}_{i}^{n}-{v}_{i-1}^{n}}{h}\right)+h{r}_{i,n}^{\left(2\right)},$ (2.13)

$h|{r}_{i,n}^{\left(2\right)}|\le C{h}^{3};$ (2.14)

$\begin{array}{l}{f}_{i+\frac{1}{2},n}{v}_{i+\frac{1}{2}}^{n}-{f}_{i-\frac{1}{2},n}{v}_{i-\frac{1}{2}}^{n}\\ ={f}_{i+\frac{1}{2},n}\frac{{v}_{i}^{n}+{v}_{i+1}^{n}}{2}-{f}_{i-\frac{1}{2},n}\frac{{v}_{i}^{n}+{v}_{i-1}^{n}}{2}-h{r}_{i,n}^{\left(3\right)},\end{array}$ (2.15)

$h|{r}_{i,n}^{\left(3\right)}|\le C{h}^{3}.$ (2.16)

$\begin{array}{l}h\Delta {t}^{-\alpha }\left[{\sum }_{j=0}^{n-1}\left({\omega }_{n-j}^{\alpha -1}-{\omega }_{n-j-1}^{\alpha -1}\right){v}_{i}^{j}+{\omega }_{0}^{\alpha -1}{v}_{i}^{n}\right]-h{r}_{i,n}^{\left(1\right)}\\ ={k}_{\alpha }\left(\frac{{v}_{i+1}^{n}-{v}_{i}^{n}}{h}-\frac{{v}_{i}^{n}-{v}_{i-1}^{n}}{h}\right)+h{r}_{i,n}^{\left(2\right)}-\left({f}_{i+\frac{1}{2},n}\frac{{v}_{i}^{n}+{v}_{i+1}^{n}}{2}-{f}_{i-\frac{1}{2},n}\frac{{v}_{i}^{n}+{v}_{i-1}^{n}}{2}\right)+h{r}_{i,n}^{\left(3\right)},\end{array}$ (2.17)

${V}_{i}^{n}$ 近似 ${v}_{i}^{n}$ ，由(2.17)式我们可以得到如下的有限体积法(FV)： $i=1,2,\cdots ,N;n=1,2,\cdots ,L$

$\begin{array}{l}h\Delta {t}^{-\alpha }\left[{\sum }_{j=0}^{n-1}\left({\omega }_{n-j}^{\alpha -1}-{\omega }_{n-j-1}^{\alpha -1}\right){V}_{i}^{j}+{\omega }_{0}^{\alpha -1}{V}_{i}^{n}\right]\\ ={k}_{\alpha }\left(\frac{{V}_{i+1}^{n}-{V}_{i}^{n}}{h}-\frac{{V}_{i}^{n}-{V}_{i-1}^{n}}{h}\right)-\left({f}_{i+\frac{1}{2},n}\frac{{V}_{i}^{n}+{V}_{i+1}^{n}}{2}-{f}_{i-\frac{1}{2},n}\frac{{V}_{i}^{n}+{V}_{i-1}^{n}}{2}\right),\end{array}$ (2.18)

$\begin{array}{l}{V}_{i}^{0}=\phi \left({x}_{i}\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}{V}_{0}^{n}={G}_{1}\left({t}_{n}\right)=\Delta {t}^{\alpha -1}{\sum }_{j=0}^{n}{\omega }_{n-j}^{1-\alpha }{g}_{1}\left({t}_{n}\right),\\ {V}_{N+1}^{n}={G}_{2}\left({t}_{n}\right)=\Delta {t}^{\alpha -1}{\sum }_{j=0}^{n}{\omega }_{n-j}^{1-\alpha }{g}_{2}\left({t}_{n}\right).\end{array}$ (2.19)

3. 数值实验

$\frac{\partial u}{\partial t}=\left(\frac{{\partial }^{2}}{\partial {x}^{2}}-\frac{\partial }{\partial x}f\left(x,t\right)\right){D}_{t}^{1-\alpha }u+g\left(x,t\right)，\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\le x\le 1,\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\le t\le 1,$ (3.1)

$\begin{array}{c}g\left(x,t\right)=2t\mathrm{cos}\left(\text{π}x\right)+\frac{2{\text{π}}^{2}}{\Gamma \left(2+\alpha \right)}{t}^{1+\alpha }\mathrm{cos}\left(\text{π}x\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}+\frac{2}{\Gamma \left(2+\alpha \right)}{t}^{1+\alpha }\left[\left(1-2x+t\right)\mathrm{cos}\left(\text{π}x\right)-f\left(x,t\right)\text{π}\mathrm{sin}\left(\text{π}x\right)\right],\end{array}$ (3.2)

Table 1. Convergence rate for space with α = 0.2, L = 50,000

Table 2. Convergence rate for space with α = 0.5, L = 50,000

Table 3. Convergence rate for space with α = 0.8, L = 50,000

Table 4. Convergence rate for time with α = 0.2, N = 5000

Table 5. Convergence rate for time α = 0.5, N = 5000

Table 6. Convergence rate for time with α = 0.8, N = 5000

$f\left(x,t\right)=x-{x}^{2}+t+xt$ 。此方程的精确解为 $u\left(x,t\right)={t}^{2}\mathrm{cos}\left(\text{π}x\right)$

$空间收敛阶=|\frac{\mathrm{ln}\left({‖细网格误差‖}_{1}/{‖粗网格误差‖}_{1}\right)}{\mathrm{ln}\left(细网格划分数N+1/粗网格划分数N+1\right)}|,$

$时间收敛阶=|\frac{\mathrm{ln}\left({‖细网格误差‖}_{1}/{‖粗网格误差‖}_{1}\right)}{\mathrm{ln}\left(细网格划分数L/粗网格划分数L\right)}|.$

A Finite Volume Method for Time Fractional Fokker-Planck Equations[J]. 应用数学进展, 2019, 08(02): 203-209. https://doi.org/10.12677/AAM.2019.82023

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