# 一类奇异摄动抛物型反应扩散问题的数值解Numerical Solution of a Class of Singularly Perturbed Parabolic Reaction-Diffusion Problem

DOI: 10.12677/AAM.2019.86136, PDF, HTML, XML, 下载: 320  浏览: 408

Abstract: The singularly perturbed problem’s high accuracy numerical method is always needed. In this paper, barycentric Lagrange interpolation collocation method (BLICM) is proposed for solving a class of singularly perturbed parabolic reaction-diffusion problem. Compared with other methods, the numerical experiment shows BLICM is a high precision method to solve this class of problems.

1. 引言

$\left\{\begin{array}{l}\frac{\partial u}{\partial t}-{\epsilon }_{1}\frac{{\partial }^{2}u}{\partial {x}^{2}}+{\alpha }_{1}u+{\alpha }_{2}v+{\alpha }_{3}r={f}_{1}\left(x,t\right),\\ \frac{\partial v}{\partial t}-{\epsilon }_{2}\frac{{\partial }^{2}v}{\partial {x}^{2}}+{\beta }_{1}u+{\beta }_{2}v+{\beta }_{3}r={f}_{2}\left(x,t\right),\left(x,t\right)\in \left[0,1\right]×\left[0,1\right],\\ \frac{\partial r}{\partial t}-{\epsilon }_{3}\frac{{\partial }^{2}r}{\partial {x}^{2}}+{\gamma }_{1}u+{\gamma }_{2}v+{\gamma }_{3}r={f}_{3}\left(x,t\right),\end{array}$ (1)

$\begin{array}{l}u\left(x,0\right)={f}_{0}\left(x\right),v\left(x,0\right)={g}_{0}\left(x\right),r\left(x,0\right)={s}_{0}\left(x\right),x\in \left[0,1\right]\\ u\left(0,t\right)={f}_{1}\left(t\right),u\left(1,t\right)={f}_{2}\left(t\right),\\ v\left(0,t\right)={g}_{1}\left(t\right),v\left(1,t\right)={g}_{2}\left(t\right),t\in \left[0,1\right]\\ r\left(0,t\right)={s}_{1}\left(t\right),r\left(1,t\right)={s}_{2}\left(t\right),\end{array}$ (2)

2. BLICM的描述

$\begin{array}{l}{x}_{i}=-\mathrm{cos}\left(\frac{i-1}{M-1}\right)\text{π},i=1,2,\cdots ,M,\\ {t}_{j}=-\mathrm{cos}\left(\frac{j-1}{N-1}\right)\text{π},j=1,2,\cdots ,N.\end{array}$ (3)

$\begin{array}{l}u\left(x,t\right)=\underset{i=1}{\overset{M}{\sum }}\underset{j=1}{\overset{N}{\sum }}{\xi }_{i}\left(x\right){\eta }_{j}\left(t\right)u\left({x}_{i},{t}_{j}\right),\\ v\left(x,t\right)=\underset{i=1}{\overset{M}{\sum }}\underset{j=1}{\overset{N}{\sum }}{\xi }_{i}\left(x\right){\eta }_{j}\left(t\right)v\left({x}_{i},{t}_{j}\right),\\ r\left(x,t\right)=\underset{i=1}{\overset{M}{\sum }}\underset{j=1}{\overset{N}{\sum }}{\xi }_{i}\left(x\right){\eta }_{j}\left(t\right)r\left({x}_{i},{t}_{j}\right),\end{array}$ (4)

$\begin{array}{l}\frac{{\partial }^{l+k}u}{\partial {x}^{l}\partial {t}^{k}}=\underset{i=1}{\overset{M}{\sum }}\underset{j=1}{\overset{N}{\sum }}{\xi }_{i}^{\left(l\right)}\left(x\right){\eta }_{j}^{\left(k\right)}\left(t\right)u\left({x}_{i},{t}_{j}\right),l,k=1,2,\cdots ,\\ \frac{{\partial }^{l+k}v}{\partial {x}^{l}\partial {t}^{k}}=\underset{i=1}{\overset{M}{\sum }}\underset{j=1}{\overset{N}{\sum }}{\xi }_{i}^{\left(l\right)}\left(x\right){\eta }_{j}^{\left(k\right)}\left(t\right)v\left({x}_{i},{t}_{j}\right),l,k=1,2,\cdots ,\\ \frac{{\partial }^{l+k}r}{\partial {x}^{l}\partial {t}^{k}}=\underset{i=1}{\overset{M}{\sum }}\underset{j=1}{\overset{N}{\sum }}{\xi }_{i}^{\left(l\right)}\left(x\right){\eta }_{j}^{\left(k\right)}\left(t\right)r\left({x}_{i},{t}_{j}\right),l,k=1,2,\cdots ,\end{array}$ (5)

$\begin{array}{l}{u}^{\left(l,k\right)}\left({x}_{p},{t}_{q}\right)=\frac{{\partial }^{l+k}u\left({x}_{p},{t}_{q}\right)}{\partial {x}^{l}\partial {t}^{k}}=\underset{i=1}{\overset{M}{\sum }}\underset{j=1}{\overset{N}{\sum }}{\xi }_{i}^{\left(l\right)}\left({x}_{p}\right){\eta }_{j}^{\left(k\right)}\left({t}_{q}\right)u\left({x}_{p},{t}_{q}\right),p=1,2,\cdots ,M;q=1,2,\cdots ,N,\\ {v}^{\left(l,k\right)}\left({x}_{p},{t}_{q}\right)=\frac{{\partial }^{l+k}v\left({x}_{p},{t}_{q}\right)}{\partial {x}^{l}\partial {t}^{k}}=\underset{i=1}{\overset{M}{\sum }}\underset{j=1}{\overset{N}{\sum }}{\xi }_{i}^{\left(l\right)}\left({x}_{p}\right){\eta }_{j}^{\left(k\right)}\left({t}_{q}\right)v\left({x}_{p},{t}_{q}\right),p=1,2,\cdots ,M;q=1,2,\cdots ,N,\\ {r}^{\left(l,k\right)}\left({x}_{p},t{}_{q}\right)=\frac{{\partial }^{l+k}r\left({x}_{p},{t}_{q}\right)}{\partial {x}^{l}\partial {t}^{k}}=\underset{i=1}{\overset{M}{\sum }}\underset{j=1}{\overset{N}{\sum }}{\xi }_{i}^{\left(l\right)}\left({x}_{p}\right){\eta }_{j}^{\left(k\right)}\left({t}_{q}\right)r\left({x}_{p},{t}_{q}\right),p=1,2,\cdots ,M;q=1,2,\cdots ,N.\end{array}$ (6)

${x}^{0}={\left[{x}_{1},{x}_{2},\cdots ,{x}_{M}\right]}^{\text{T}}$${t}^{0}={\left[{t}_{1},{t}_{2},\cdots ,{t}_{N}\right]}^{\text{T}}$ 分别定义为x轴和t轴，由张量型节点坐标组成的矩阵X和T

$X={\left[{\left({x}^{0}\right)}^{\text{T}},{\left({x}^{0}\right)}^{\text{T}},\cdots ,{\left({x}^{0}\right)}^{\text{T}}\right]}^{\text{T}},T=\left[{t}^{0},{t}^{0},\cdots ,{t}^{0}\right].$ (7)

$x,t$ 是由矩阵 $X,T$ 拉伸的 $\left(N×M\right)$ 维列向量：

$x={\left[{X}_{1},{X}_{2},\cdots ,{X}_{M×N}\right]}^{\text{T}},\text{\hspace{0.17em}}t={\left[{T}_{1},{T}_{2},\cdots ,{T}_{M×N}\right]}^{\text{T}}.$ (8)

${X}_{k}={X}_{\left(i-1\right)N+j}={x}_{i},\text{\hspace{0.17em}}{T}_{k}={T}_{\left(i-1\right)N+j}={t}_{j},\text{\hspace{0.17em}}i=1,2,\cdots ,M;J=1,2,\cdots ,N;k=1,2,\cdots ,M×N.$ (9)

$u={\left[{u}_{1},{u}_{2},\cdots ,{u}_{M×N}\right]}^{\text{T}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}{u}^{\left(l,k\right)}={\left[{u}_{1}^{\left(l,k\right)},{u}_{2}^{\left(l,k\right)},\cdots ,{u}_{M×N}^{\left(l,k\right)}\right]}^{\text{T}},$

${u}_{p}=u\left({X}_{p},{T}_{p}\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}{u}_{p}^{\left(l,k\right)}={u}^{\left(l,k\right)}\left({X}_{p},{T}_{p}\right),\text{\hspace{0.17em}}p=1,2,\cdots ,M×N.$

$v={\left[{v}_{1},{v}_{2},\cdots ,{v}_{M×N}\right]}^{\text{T}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}{v}^{\left(l,k\right)}={\left[{v}_{1}^{\left(l,k\right)},{v}_{2}^{\left(l,k\right)},\cdots ,{v}_{M×N}^{\left(l,k\right)}\right]}^{\text{T}},$

${v}_{p}=v\left({X}_{p},{T}_{p}\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}{v}_{p}^{\left(l,k\right)}={v}^{\left(l,k\right)}\left({X}_{p},{T}_{p}\right),\text{\hspace{0.17em}}p=1,2,\cdots ,M×N.$

$r={\left[{r}_{1},{r}_{2},\cdots ,{r}_{M×N}\right]}^{\text{T}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}{r}_{p}^{\left(l,k\right)}={\left[{r}_{1}^{\left(l,k\right)},{r}_{2}^{\left(l,k\right)},\cdots ,{r}_{M×N}^{\left(l,k\right)}\right]}^{\text{T}},$

${r}_{p}=r\left({X}_{p},{T}_{p}\right),\text{\hspace{0.17em}}\text{\hspace{0.17em}}{r}_{p}^{\left(l,k\right)}={r}^{\left(l,k\right)}\left({X}_{p},{T}_{p}\right),\text{\hspace{0.17em}}p=1,2,\cdots ,M×N.$

${u}^{\left(l,k\right)}={D}^{\left(l,k\right)}u,\text{\hspace{0.17em}}\text{\hspace{0.17em}}{v}^{\left(l,k\right)}={D}^{\left(l,k\right)}v,\text{\hspace{0.17em}}\text{\hspace{0.17em}}{r}^{\left(l,k\right)}={D}^{\left(l,k\right)}r.$ (10)

$\left[\begin{array}{ccc}{D}^{\left(0,1\right)}-{\epsilon }_{1}{D}^{\left(2,0\right)}+{\alpha }_{1}I& {\alpha }_{2}I& {\alpha }_{3}I\\ {\beta }_{1}I& {D}^{\left(0,1\right)}-{\epsilon }_{2}{D}^{\left(2,0\right)}+{\beta }_{2}I& {\beta }_{3}I\\ {\gamma }_{1}I& {\gamma }_{2}I& {D}^{\left(0,1\right)}-{\epsilon }_{3}{D}^{\left(2,0\right)}+{\gamma }_{3}I\end{array}\right]\left[\begin{array}{l}u\\ v\\ r\end{array}\right]=\left[\begin{array}{l}{f}_{1}\\ {f}_{2}\\ {f}_{3}\end{array}\right]$ (11)

$\begin{array}{l}\underset{i=1}{\overset{M}{\sum }}\underset{j=1}{\overset{N}{\sum }}{\xi }_{i}\left({x}_{p}\right){\eta }_{j}\left(0\right){u}_{ij}={f}_{0}\left({x}_{p}\right),\\ \underset{i=1}{\overset{M}{\sum }}\underset{j=1}{\overset{N}{\sum }}{{\xi }^{\prime }}_{i}\left({x}_{p}\right){{\eta }^{\prime }}_{j}\left(0\right){v}_{ij}={g}_{0}\left({x}_{p}\right),\\ \underset{i=1}{\overset{M}{\sum }}\underset{j=1}{\overset{N}{\sum }}{{\xi }^{″}}_{i}\left({x}_{p}\right){{\eta }^{″}}_{j}\left(0\right){r}_{ij}={s}_{0}\left({x}_{p}\right),\end{array}$ $\begin{array}{l}\underset{i=1}{\overset{M}{\sum }}\underset{j=1}{\overset{N}{\sum }}{\xi }_{i}\left(0\right){\eta }_{j}\left({t}_{q}\right){u}_{ij}={f}_{1}\left({t}_{q}\right),\\ \underset{i=1}{\overset{M}{\sum }}\underset{j=1}{\overset{N}{\sum }}{{\xi }^{\prime }}_{i}\left(0\right){{\eta }^{\prime }}_{j}\left({t}_{q}\right){v}_{ij}={g}_{1}\left({t}_{q}\right),\\ \underset{i=1}{\overset{M}{\sum }}\underset{j=1}{\overset{N}{\sum }}{{\xi }^{″}}_{i}\left(0\right){{\eta }^{″}}_{j}\left({t}_{q}\right){r}_{ij}={s}_{1}\left({t}_{q}\right),\end{array}$ $\begin{array}{l}\underset{i=1}{\overset{M}{\sum }}\underset{j=1}{\overset{N}{\sum }}{\xi }_{i}\left(1\right){\eta }_{j}\left({t}_{q}\right){u}_{ij}={f}_{2}\left({t}_{q}\right),\\ \underset{i=1}{\overset{M}{\sum }}\underset{j=1}{\overset{N}{\sum }}{\xi }_{i}{}^{\prime }\left(1\right){{\eta }^{\prime }}_{j}\left({t}_{q}\right){v}_{ij}={g}_{2}\left({t}_{q}\right),\\ \underset{i=1}{\overset{M}{\sum }}\underset{j=1}{\overset{N}{\sum }}{{\xi }^{″}}_{i}\left(1\right){{\eta }^{″}}_{j}\left({t}_{q}\right){r}_{ij}={s}_{2}\left({t}_{q}\right),\end{array}$ (12)

3. 数值算例

$\left\{\begin{array}{l}\frac{\partial u}{\partial t}-{\epsilon }_{1}\frac{{\partial }^{2}u}{\partial {x}^{2}}+5u-tv={f}_{1}\left(x,t\right),\\ \frac{\partial v}{\partial t}-{\epsilon }_{2}\frac{{\partial }^{2}v}{\partial {x}^{2}}-tu+5\left(1+xt\right)v-r={f}_{2}\left(x,t\right),\\ \frac{\partial r}{\partial t}-{\epsilon }_{3}\frac{{\partial }^{2}r}{\partial {x}^{2}}-2u-\left(1+{x}^{2}\right)v+\left(7+xt\right)r={f}_{3}\left(x,t\right),\end{array}$ (13)

$u\left(0,t\right)=v\left(0,t\right)=r\left(0,t\right)=0,u\left(1,t\right)=v\left(1,t\right)=r\left(1,t\right)=0,u\left(x,0\right)=v\left(x,0\right)=r\left(x,0\right)=0,$ (14)

${f}_{1}\left(x,t\right)=\left\{\begin{array}{l}2+x+t,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\le x\le 0.5,\\ 1,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }0.5

${f}_{2}\left(x,t\right)=\left\{\begin{array}{l}2\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }0\le x\le 0.5,\\ t,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0.5 (15)

${f}_{3}\left(x,t\right)=\left\{\begin{array}{l}1+x,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{ }\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\le x\le 0.5,\\ {x}^{2}+t,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0.5

Figure 1. Surface plot of the Example 1 for ${\epsilon }_{1}={10}^{-9},{\epsilon }_{2}={10}^{-5},{\epsilon }_{3}={10}^{-3}$ and $N=256,M=8$

Figure 2. Surface plot of the Example 1 for ${\epsilon }_{1}={10}^{-9},{\epsilon }_{2}={10}^{-5},{\epsilon }_{3}={10}^{-3}$ and $N=128,M=2$

Table 1. Comparison of maximum point-wise errors and ε 1 , ε 2 , ε 3 uniform rate of convergence for the example 1

Table 2. Comparison of maximum point-wise errors and ε 1 , ε 2 , ε 3 uniform rate of convergence for the Example 1

Table 3. Number of iterations required by the algorithm for Example 1

$\left\{\begin{array}{l}\frac{\partial u}{\partial t}-{\epsilon }_{1}\frac{{\partial }^{2}u}{\partial {x}^{2}}+2.1u-\left(1-x\right)v-\left(1+x\right)r=16{x}^{2}{\left(1-x\right)}^{2},\\ \frac{\partial v}{\partial t}-{\epsilon }_{2}\frac{{\partial }^{2}v}{\partial {x}^{2}}-xu+\left(1.1+x\right)v-xr={t}^{2},\\ \frac{\partial r}{\partial t}-{\epsilon }_{3}\frac{{\partial }^{2}r}{\partial {x}^{2}}-\left(2+x\right)u-\left(1-x\right)v-\left(3.1+x\right)r=-16{x}^{2}{\left(1-x\right)}^{2},\end{array}$ (16)

$u\left(0,t\right)=v\left(0,t\right)=r\left(0,t\right)=0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}u\left(1,t\right)=v\left(1,t\right)=r\left(1,t\right)=0.$

Figure 3. Numerical solution of Example 2 for ${\epsilon }_{1}={10}^{-1},{\epsilon }_{2}={10}^{-2},{\epsilon }_{3}={10}^{-3}$ and $N=64,M=8$

Table 4. Comparison of maximum point-wise errors and convergence rate for the Example 2

$\left\{\begin{array}{l}\frac{\partial u}{\partial t}-{\epsilon }_{1}\frac{{\partial }^{2}u}{\partial {x}^{2}}+\left(5+xt\right)u-\left({x}^{2}+{t}^{2}\right)v-\left(2+xt\right)r=2\left(1-{\text{e}}^{-t}\right)+5t\mathrm{cos}\left(xt\right),\\ \frac{\partial v}{\partial t}-{\epsilon }_{2}\frac{{\partial }^{2}v}{\partial {x}^{2}}-5xt{\text{e}}^{-x}u+\left(6+\mathrm{cos}\left(xt\right)\right)v-\mathrm{cos}\left(xt\right)r=\left(10x+1\right)t\left({x}^{2}+{t}^{2}-5\mathrm{sin}\left(xt\right)\right),\\ \frac{\partial r}{\partial t}-{\epsilon }_{3}\frac{{\partial }^{2}r}{\partial {x}^{2}}-3x{\text{e}}^{-t}u-\left(xt+\mathrm{sin}\left(x+t\right)\right)v+\left(5+\left(x+1\right){\text{e}}^{-xt}\right)r=-3t\mathrm{cos}\left(x+t\right).\\ u\left(0,t\right)=u\left(1,t\right)=0,t\in \left[0,1\right],u\left(x,0\right)=0,x\in \left[0,1\right].\end{array}$ (17)

Figure 4. The numerical solution for Example 3 at ${\epsilon }_{1}={10}^{-8},{\epsilon }_{2}={10}^{-6},{\epsilon }_{3}={10}^{-4}$

Table 5. Comparison of maximum errors and orders of convergence of Example 3

Table 6. Comparison of CPU time for Example 3

$\left\{\begin{array}{l}\frac{\partial u}{\partial t}-{\epsilon }_{1}\frac{{\partial }^{2}u}{\partial {x}^{2}}+\left(1+0.5xy\right)\frac{\partial u}{\partial x}+{\text{e}}^{{x}^{2}y}\frac{\partial u}{\partial y}+{\text{e}}^{x+y}\left(1+t\right)u-t\left(x+y\right)v-txr=10{t}^{2}\mathrm{sin}\left(x+y\right),\\ \frac{\partial v}{\partial t}-{\epsilon }_{2}\frac{{\partial }^{2}v}{\partial {x}^{2}}+\left(5+{x}^{2}y\right)\frac{\partial u}{\partial x}+\left(3+\mathrm{sin}\left(x+y\right)\right)\frac{\partial u}{\partial y}-\left(x+y\right)u+\left(1+t\right)\left(3+x+y\right)v-t\mathrm{sin}\left(y\right)r=-5\left(1-{\text{e}}^{-t}\right)\left({x}^{2}+{y}^{2}\right),\\ \frac{\partial r}{\partial t}-{\epsilon }_{3}\frac{{\partial }^{2}r}{\partial {x}^{2}}+\left(3-xy\right)\frac{\partial u}{\partial x}+\left(1+x+y\right)\frac{\partial u}{\partial y}-x{y}^{2}u-t\left(\mathrm{sin}x+\mathrm{sin}y\right)v+{\text{e}}^{t}\left(2+\mathrm{cos}\left(x+y\right)\right)r=-4t{\text{e}}^{t}\mathrm{cos}\left(xy\right),\end{array}$

$\begin{array}{l}u\left(0,y,t\right)=4y\mathrm{sin}\left(t\right),\\ u\left(x,0,t\right)=4x\mathrm{sin}\left(t\right),\end{array}$ $\begin{array}{l}v\left(0,y,t\right)=0,\\ v\left(x,0,t\right)=0,\end{array}$ $\begin{array}{l}r\left(0,y,t\right)=3\left(1-{\text{e}}^{t}\right),\\ r\left(x,0,t\right)=3\left(1-{\text{e}}^{t}\right),\end{array}$

$u\left(x,y,0\right)=v\left(x,y,0\right)=r\left(x,y,0\right)=0,$

$\begin{array}{l}u\left(1,y,t\right)=4\left(1+y\right)\mathrm{sin}\left(t\right),\\ u\left(x,1,t\right)=4\left(x+1\right)\mathrm{sin}\left(t\right),\end{array}$ $\begin{array}{l}v\left(1,y,t\right)=y{t}^{2},\\ v\left(x,1,t\right)=x{t}^{2},\end{array}$ $\begin{array}{l}r\left(1,y,t\right)=3{\text{e}}^{y}\left(1-{\text{e}}^{t}\right),\\ r\left(x,1,t\right)=3{\text{e}}^{x}\left(1-{\text{e}}^{t}\right).\end{array}$

Figure 5. Numerical solution of Example 4 at T = 1 for $\epsilon ={10}^{-4}$ and $N=32,M=32$

Table 7. Comparison of maximum errors and orders of convergence in Example 4 for v

4. 结论

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