Burgers方程混合问题的Lagrange插值逼近

Lagrange Interpolation Approximation of Mixed Problem of Burgers Equation

摘要: 利用等距节点为插值节点，构造Burgers方程混合问题的时空二元Lagrange插值逼近格式。即在时间和空间方向都采用Lagrange插值多项式进行逼近，化为非线性方程组，利用迭代方法进行求解。最后通过数值结果证明了算法内容的正确性与实用性，为研究其他问题提供了强有力的工具。

Abstract: The spatiotemporal binary Lagrange interpolation approximation scheme for the mixed problem of Burgers equation is constructed by using equidistant nodes as interpolation nodes. That is, La-grange interpolation polynomials are used to approximate in time and space, which are trans-formed into nonlinear equations and solved by fixed point iterative method. Finally, the numerical results prove the correctness and practicability of the algorithm, which provides a powerful tool for studying other problems.

1. 引言

Burgers方程 [1] 作为一种非线性偏微分方程，此方程在工程上有着重要的应用，它可以用来描述湍流，车队的交通流，氏族的随机迁移，化学工程中的分离等现象。

因此，对Burgers方程的研究及对Burgers方程的精确解 [2] 的分析都有着重要的理论意义和应用价值。有学者采用最小二乘混合有限元方法 [3]，有限差分法 [4]，有限体积法 [5] 等方法研究。另外谱方法可以有效求解偏微分方程，这种数值方法由于高精度而被广泛应用。因此有学者采用Jacobi多项式谱方法，时空耦合谱元方法和时空Chebyshev伪谱方法等方法针对Burgers方程的混合问题 [6] [7] [8] 进行求解。根据众多学者的研究，由于Lagrange插值 [9] 的简单有效，而被广泛运用到此方程的求解中。

最近，文献 [10] 考虑了Korteweg-de Vries方程的Legendre-Hermite时空二元谱配置方法，是非等距节点配置方法。更有兴趣的问题是如何使用等距节点为配置点构造二元插值多项式逼近非线性偏微分方程的解。因此作为一种有益的探索和尝试，本文将构造Burgers方程的混合问题的时空等距节点的二元Lagrange插值逼近方法。

记 $\frac{\partial }{{\partial }_x}$ 为 ${\partial }_x$。令 $u\left(x,t\right)$ 是实值函数， $x\in \left(a,b\right)$， $t\in \left(0,T\right]$。考虑Burgers方程混合问题：

$\{\begin{array}{l}{\partial }_{t}u+u{\partial }_{x}u-{\partial }_x^{2}u=0,x\in \left(a,b\right),t\in \left(0,T\right],\\ u\left(a,t\right)=g_1\left(t\right),u\left(b,t\right)=g_2\left(t\right),t\in \left[0,T\right],\\ u\left(x,0\right)=\phi \left(x\right),x\in \left[a,b\right].\end{array}$ (1)

采用时空等距节点的二元Lagrange插值逼近方法来逼近(1)的精确解，即在时间和空间方向均构造等距节点的Lagrange插值逼近格式，采用不动点迭代方法对(1)式所推导出非线性代数方程组求取近似解。分析可知，所构造的算法格式在对时空方向进行插值逼近时，简单明了，易于计算，从而便于理解与实际应用，而且当函数充分光滑时，逼近效果良好，精度高。

2. Burgers方程的数值求解

2.1. 基于等距节点Lagrange插值多项式的微分矩阵

记 $h=\left(b-a\right)/N,x_0=a,x_j=x_0+kh,k=0,1,\cdots ,N$。则与 $x_k$ 为节点的拉格朗日插值基函数为：

${\varphi }_m\left(x\right)=\frac{\left(x-x_0\right)\left(x-x_1\right)\cdots \left(x-x_{m-1}\right)\left(x-x_{m+1}\right)\cdots \left(x-x_N\right)}{\left(x_m-x_0\right)\left(x_m-x_1\right)\cdots \left(x_m-x_{m-1}\right)\left(x_m-x_{m+1}\right)\cdots \left(x_m-x_N\right)},m=0,1,\cdots ,N.$

令 $\omega \left(x\right)=\left(x-x_0\right)\left(x-x_1\right)\cdots \left(x-x_N\right)$，有：

${\varphi }_m\left(x\right)=\frac{\omega \left(x\right)}{\left(x-x_m\right){\partial }_x\omega \left(x_m\right)}.$ (2)

记 $P_N\left(a,b\right)$ 为次数 $\le N$ 的多项式集合，对于 $u\left(x\right)\in C\left(a,b\right)$，其插值多项式为：

$p_N\left(x\right)={\displaystyle \underset{m=0}{\overset{N}{\sum }}{u}_m{\varphi }_m}\left(x\right),u_m=u\left(x_m\right),x\in \left[a,b\right],$

对 $p_N\left(x\right)$ 关于x求一阶导数，并令 $x=x_k,k=0,1,2,\cdots ,N$ 得：

${\partial }_x{p}_N\left(x_k\right)={\displaystyle \underset{m=0}{\overset{N}{\sum }}{\partial }_x{\varphi }_m\left(x_k\right)u_m}.$

引理1设 $D=\left(d_{km}\right)$ 是 $\left(N+1\right)\times \left(N+1\right)$ 矩阵，且 $d_{km}={\partial }_x{\varphi }_m\left(x_k\right)$，则有：

$d_{km}=\{\begin{array}{l}\frac{{\left(-1\right)}^{m-k}k!\left(N-k\right)!}{h\left(k-m\right)m!\left(N-m\right)!},k\ne m,\\ \frac{1}{h}\left(\left(1+\frac{1}{2}+\cdots +\frac{1}{m}\right)-\left(1+\frac{1}{2}+\cdots +\frac{1}{N-m}\right)\right),0<k=m<N,\\ -\frac{1}{h}\left(1+\frac{1}{2}+\cdots +\frac{1}{N}\right),k=m=0,\\ \frac{1}{h}\left(1+\frac{1}{2}+\cdots +\frac{1}{N}\right),k=m=N.\end{array}$ (3)

进一步，再对 $p_N\left(x\right)$ 求二阶导数，令 $x=x_k$，得：

${\partial }_x^2{p}_N\left(x_k\right)={\displaystyle \underset{m=0}{\overset{N}{\sum }}{d}_{km}^{\left(2\right)}{u}_m}\begin{array}{cc},& d_{km}^{\left(2\right)}={\partial }_x^2{\varphi }_m\end{array}\left(x_k\right).$ (4)

这里 $\hat{D}=\left({\hat{D}}_{km}\right)=D^2$。

证明：对基函数求导得：

${\partial }_x{\varphi }_m\left(x\right)=\frac{{\omega }^{\prime }\left(x\right)\left(x-x_m\right)-\omega \left(x\right)}{{\left(x-x_m\right)}^2{\omega }^{\prime }\left(x_m\right)}.$

令 $x=x_k,k\ne m$，可得：

${\partial }_x{\varphi }_m\left(x_k\right)=\frac{{\omega }^{\prime }\left(x_k\right)}{\left(x_k-x_m\right){\omega }^{\prime }\left(x_m\right)}.$

令 $x\to x_m$，利用洛必达法则得：

${\partial }_x{\varphi }_m\left(x_m\right)=\frac{{\partial }_x\omega \left(x_m\right)}{2{\partial }_x^2\omega \left(x_m\right)}.$

所有：

${\partial }_x{\varphi }_m\left(x_k\right)=\{\begin{array}{l}\frac{{\partial }_x\omega \left(x_k\right)}{\left(x_k-x_m\right){\partial }_x\omega \left(x_m\right)},k\ne m,\\ \frac{{\partial }_x^2\omega \left(x_m\right)}{2{\partial }_x\omega \left(x_m\right)},k=m.\end{array}$ (5)

直接计算有：

${\partial }_x\omega \left(x_k\right)=k!\left(N-k\right)!{\left(-1\right)}^{N-k}{h}^N,h=\left(b-a\right)/N.$

令 ${\phi }_m\left(x\right)=\left(x-x_0\right)\left(x-x_1\right)\cdots \left(x-x_{m-1}\right)\left(x-x_{m+1}\right)\cdots \left(x-x_N\right)$，可得：

$\left(x-x_m\right){\phi }_m\left(x\right)=\omega \left(x\right).$

对上式求导两次并 $x=x_m$，有：

${\partial }_x^2\omega \left(x_m\right)=2{\partial }_x{\phi }_m\left(x_m\right).$

直接计算有：

${\partial }_x{\phi }_m\left(x_m\right)=\frac{1}{h}\left(\left(1+\frac{1}{2}+\cdots +\frac{1}{m}\right)-\left(1+\frac{1}{2}+\cdots +\frac{1}{N-m}\right)\right){\phi }_m\left(x_m\right),m\ne 0,N,{\phi }_m\left(x_m\right)={\partial }_x\omega \left(x_m\right).$.

特别地，

${\partial }_x{\phi }_0\left(x_0\right)=-\frac{1}{h}\left(1+\frac{1}{2}+\cdots +\frac{1}{N}\right){\phi }_0\left(x_0\right),\begin{array}{cc}& {\partial }_x{\phi }_N\left(x_N\right)=\frac{1}{h}\left(1+\frac{1}{2}+\cdots +\frac{1}{N}\right){\phi }_N\left(x_N\right).\end{array}$

上述结果代入(5)式，(3)式得证。

仿照文献 [11] 中的(3.71)式，可证得 $\hat{D}=D^2$。引理1得证。

实际上，若 $D^{\left(j\right)}$ 表示j 阶微分矩阵，我们有： $D^{\left(j\right)}=D^j$，证明见文献 [11]。

特别地，用 ${\psi }_m\left(t\right)$ 表示时间方向的插值多项式基函数，记 ${\hat{d}}_{lm}={\partial }_t{\psi }_m\left(t_l\right)$， $t_l$ 为时间方向插值节点。可以推导类似于(3)式的微分矩阵元素表达式，只需令 $a=0,b=T,h=T/M$ 即可。我们略去具体推导过程。

2.2. Burgers方程混合问题的Lagrange插值逼近算法格式

令 $\Omega =\left(a,b\right)\otimes \left(0,T\right),P_{M,N}\left(\Omega \right)=P_M\left(0,T\right)\otimes P_N\left(a,b\right)$。(1)式的时空二元插值逼近方法就是求多项式 $u_{M,N}\left(x,t\right)\in P_{M,N}\left(\Omega \right)$ 满足：

$\{\begin{array}{l}{\partial }_t{u}_{M,N}+u{\partial }_x{u}_{M,N}-{\partial }_x^2{u}_{M,N}=0,x\in \left(a,b\right),t\in \left(0,T\right],\\ u_{M,N}\left(a,t\right)=g_1\left(t\right),u_{M,N}\left(b,t\right)=g_2\left(t\right),t\in \left[0,T\right],\\ u_{M,N}\left(x,0\right)=\phi \left(x\right),x\in \left[a,b\right].\end{array}$ (6)

将数值解展开为：

$u_{M,N}\left(x,t\right)={\displaystyle \underset{n=0}{\overset{N}{\sum }}{\displaystyle \underset{m=0}{\overset{M}{\sum }}{\hat{u}}_{m,n}}}{\varphi }_n\left(x\right){\psi }_m\left(t\right).$

逼近(1)的解，将其代入(1)式可得：

$\{\begin{array}{l}{\displaystyle \underset{n=0}{\overset{N}{\sum }}{\displaystyle \underset{m=0}{\overset{M}{\sum }}{\hat{u}}_{m,n}}}{\varphi }_n\left(x_k\right){\partial }_t{\psi }_m\left(t_l\right)+{\displaystyle \underset{n=0}{\overset{N}{\sum }}{\displaystyle \underset{m=0}{\overset{M}{\sum }}{\hat{u}}_{m,n}}}{\varphi }_n\left(x_k\right){\psi }_m\left(t_l\right){\displaystyle \underset{n=0}{\overset{N}{\sum }}{\displaystyle \underset{m=0}{\overset{M}{\sum }}{\hat{u}}_{m,n}}}{\partial }_x{\varphi }_n\left(x_k\right){\psi }_m\left(t_l\right)\\ -{\displaystyle \underset{n=0}{\overset{N}{\sum }}{\displaystyle \underset{m=0}{\overset{M}{\sum }}{\hat{u}}_{m,n}}}{\partial }_x^2{\varphi }_n\left(x_k\right){\psi }_m\left(t_l\right)=0,k=1,2,\cdots ,N-1;l=1,2,\cdots ,M;\\ {\displaystyle \underset{n=0}{\overset{N}{\sum }}{\displaystyle \underset{m=0}{\overset{M}{\sum }}{\hat{u}}_{m,n}}}{\varphi }_n\left(x_0\right){\psi }_m\left(t_l\right)=g_1\left(t_l\right),{\displaystyle \underset{n=0}{\overset{N}{\sum }}{\displaystyle \underset{m=0}{\overset{M}{\sum }}{\hat{u}}_{m,n}}}{\varphi }_n\left(x_N\right){\psi }_m\left(t_l\right)=g_2\left(t_l\right),l=1,2,\cdots ,M;\\ {\hat{u}}_{0,k}=\phi \left(x_k\right),k=0,1,\cdots ,N.\end{array}$ (7)

由微分矩阵定义，(7)式等价地表示为：

$\{\begin{array}{l}{\displaystyle \underset{m=0}{\overset{M}{\sum }}{\hat{u}}_{m,k}}{\hat{d}}_{l,m}+{\hat{u}}_{l,k}{\displaystyle \underset{n=0}{\overset{N}{\sum }}{\hat{u}}_{l,n}}{d}_{k,n}-{\displaystyle \underset{n=0}{\overset{N}{\sum }}{\hat{u}}_{l,n}}{d}_{k,n}^{\left(2\right)}=0,\\ k=1,2,\cdots ,N-1;l=1,2,\cdots ,M;\\ {\hat{u}}_{l,0}=g_1\left(t_l\right),{\hat{u}}_{l,N}=g_2\left(t_l\right),l=1,2,\cdots ,M;\\ {\hat{u}}_{0,k}=\phi \left(x_k\right),k=0,1,2,\cdots ,N.\end{array}$ (8)

则(8)式为：

$\{\begin{array}{l}{\displaystyle \underset{m=1}{\overset{M}{\sum }}{\hat{u}}_{m,k}}{\hat{d}}_{l,m}+{\hat{u}}_{l,k}{\displaystyle \underset{n=1}{\overset{N-1}{\sum }}{\hat{u}}_{l,n}}{d}_{k,n}-{\displaystyle \underset{n=1}{\overset{N-1}{\sum }}{\hat{u}}_{l,n}}{d}_{k,n}^{\left(2\right)}\\ ={\hat{u}}_{l,0}{d}_{k,0}^{\left(2\right)}+{\hat{u}}_{l,N}{d}_{k,N}^{\left(2\right)}-{\hat{u}}_{0,k}{\hat{d}}_{l,0}-{\hat{u}}_{l,k}\left({\hat{u}}_{l,0}{d}_{k,0}+{\hat{u}}_{l,N}{d}_{k,N}\right),\\ k=1,2,\cdots ,N-1;l=1,2,\cdots ,M.\end{array}$ (9)

令 $l=1,2,\cdots ,M$，可得：

这里“ $.*$ ”表示矩阵对应元素相乘。

令 $k=1,2,\cdots ,N-1$，记：

$A=\left(\begin{array}{cccc}{\hat{d}}_{1,1}& {\hat{d}}_{1,2}& \cdots & {\hat{d}}_{1,M}\\ {\hat{d}}_{2,1}& {\hat{d}}_{2,2}& \cdots & {\hat{d}}_{2,M}\\ ⋮& ⋮& \ddots & ⋮\\ {\hat{d}}_{M,1}& {\hat{d}}_{M,2}& \cdots & {\hat{d}}_{M,M}\end{array}\right),B=\left(\begin{array}{cccc}{d}_{1,1}& d_{1,2}& \cdots & d_{1,N-1}\\ d_{2,1}& d_{2,2}& \cdots & d_{2,N-1}\\ ⋮& ⋮& \ddots & ⋮\\ d_{N-1,1}& d_{N-1,2}& \cdots & d_{N-1,N-1}\end{array}\right),$

$C=\left(\begin{array}{cccc}{d}_{1,2}^{\left(2\right)}& d_{1,1}^{\left(2\right)}& \cdots & d_{1,N-1}^{\left(2\right)}\\ d_{2,1}^{\left(2\right)}& d_{2,2}^{\left(2\right)}& \cdots & d_{2,N-1}^{\left(2\right)}\\ ⋮& ⋮& \ddots & ⋮\\ d_{N-1,1}^{\left(2\right)}& d_{N-1,2}^{\left(2\right)}& \cdots & d_{N-1,N-1}^{\left(2\right)}\end{array}\right),X=\left(\begin{array}{cccc}{\hat{u}}_{1,1}& {\hat{u}}_{1,2}& \cdots & {\hat{u}}_{1,N-1}\\ {\hat{u}}_{2,1}& {\hat{u}}_{2,2}& \cdots & {\hat{u}}_{2,N-1}\\ ⋮& ⋮& \ddots & ⋮\\ {\hat{u}}_{M,1}& {\hat{u}}_{M,2}& \cdots & {\hat{u}}_{M,N-1}\end{array}\right),$

$F_1=\left(\begin{array}{c}{\hat{u}}_{\text{1},0}\\ {\hat{u}}_{2,0}\\ ⋮\\ {\hat{u}}_{M,0}\end{array}\right)\left(\begin{array}{cccc}{d}_{1,0}^{\left(2\right)}& d_{2,0}^{\left(2\right)}& \cdots & d_{N-1,0}^{\left(2\right)}\end{array}\right)+\left(\begin{array}{c}{\hat{u}}_{\text{1},N}\\ {\hat{u}}_{2,N}\\ ⋮\\ {\hat{u}}_{M,N}\end{array}\right)\left(\begin{array}{cccc}{d}_{1,N}^{\left(2\right)}& d_{2,N}^{\left(2\right)}& \cdots & d_{N-1,N}^{\left(2\right)}\end{array}\right),$

$F_2=\left(\begin{array}{c}{\hat{u}}_{\text{1},0}\\ {\hat{u}}_{2,0}\\ ⋮\\ {\hat{u}}_{M,0}\end{array}\right)\left(\begin{array}{cccc}{d}_{1,0}& d_{2,0}& \cdots & d_{N-1,0}\end{array}\right)+\left(\begin{array}{c}{\hat{u}}_{\text{1},N}\\ {\hat{u}}_{2,N}\\ ⋮\\ {\hat{u}}_{M,N}\end{array}\right)\left(\begin{array}{cccc}{d}_{1,N}& d_{2,N}& \cdots & d_{N-1,N}\end{array}\right),$

$D=\left(\begin{array}{c}{\hat{d}}_{1,0}\\ {\hat{d}}_{2,0}\\ ⋮\\ {\hat{d}}_{M,0}\end{array}\right)\left(\begin{array}{cccc}{\hat{u}}_{0,0}& {\hat{u}}_{0,1}& \cdots & {\hat{u}}_{0,N-1}\end{array}\right).$

可得如下非线性矩阵方程：

$AX+X.*\left(X{B}^{\text{T}}\right)=X{C}^{\text{T}}-D-X.*F_2+F_1.$ (10)

3. 数值结果

将(10)式化为非线性方程组，选取不动点迭代方法求其近似解。令

$\begin{array}{l}Y={\left({\hat{u}}_{1,1}{\hat{u}}_{1,2}\cdots {\hat{u}}_{1,N-1},{\hat{u}}_{2,1}\cdots {\hat{u}}_{2,N-1},\cdots {\hat{u}}_{M,1}{\hat{u}}_{M,2}\cdots {\hat{u}}_{M,N-1}\right)}^{\text{T}},\\ Y_0={\left({\hat{u}}_{0,1}{\hat{u}}_{0,2}\cdots {\hat{u}}_{0,N-1},\begin{array}{cccc}0& 0& \cdots & 0,\cdots \begin{array}{cccc}0& 0& \cdots & 0\end{array}\end{array}\right)}^{\text{T}},\\ R={\left({\hat{d}}_{1,0}\left({\hat{u}}_{0,1}\begin{array}{cc}{\hat{u}}_{0,2}\cdots & {\hat{u}}_{0,N-1}\end{array}\right),{\hat{d}}_{2,0}\left({\hat{u}}_{0,1}\begin{array}{cc}{\hat{u}}_{0,2}\cdots & {\hat{u}}_{0,N-1}\end{array}\right),\cdots ,{\hat{d}}_{M,0}\left({\hat{u}}_{0,1}\begin{array}{cc}{\hat{u}}_{0,2}\cdots & {\hat{u}}_{0,N-1}\end{array}\right)\right)}^{\text{T}},\end{array}$

$\begin{array}{l}{H}_1={\left({\hat{u}}_{1,0}\left(\begin{array}{cc}{d}_{1,0}^{\left(2\right)}& d_{2,0}^{\left(2\right)}\cdots \end{array}{d}_{N-1,0}^{\left(2\right)}\right),{\hat{u}}_{2,0}\left(\begin{array}{cc}{d}_{1,0}^{\left(2\right)}& d_{2,0}^{\left(2\right)}\cdots \end{array}{d}_{N-1,0}^{\left(2\right)}\right),\cdots ,{\hat{u}}_{M,0}\left(\begin{array}{cc}{d}_{1,0}^{\left(2\right)}& d_{2,0}^{\left(2\right)}\cdots \end{array}{d}_{N-1,0}^{\left(2\right)}\right)\right)}^{\text{T}}\\ +{\left({\hat{u}}_{1,N}\left(\begin{array}{cc}{d}_{1,N}^{\left(2\right)}& d_{2,N}^{\left(2\right)}\cdots \end{array}{d}_{N-1,N}^{\left(2\right)}\right),{\hat{u}}_{2,N}\left(\begin{array}{cc}{d}_{1,N}^{\left(2\right)}& d_{2,N}^{\left(2\right)}\cdots \end{array}{d}_{N-1,N}^{\left(2\right)}\right),\cdots ,{\hat{u}}_{M,N}\left(\begin{array}{cc}{d}_{1,N}^{\left(2\right)}& d_{2,N}^{\left(2\right)}\cdots \end{array}{d}_{N-1,N}^{\left(2\right)}\right)\right)}^{\text{T}},\\ H_2={\left({\hat{u}}_{1,0}\left(\begin{array}{cc}{d}_{1,0}& d_{2,0}\cdots \end{array}{d}_{N-1,0}\right),{\hat{u}}_{2,0}\left(\begin{array}{cc}{d}_{1,0}& d_{2,0}\cdots \end{array}{d}_{N-1,0}\right),\cdots ,{\hat{u}}_{M,0}\left(\begin{array}{cc}{d}_{1,0}& d_{2,0}\cdots \end{array}{d}_{N-1,0}\right)\right)}^{\text{T}}\\ +{\left({\hat{u}}_{1,N}\left(\begin{array}{cc}{d}_{1,N}& d_{2,N}\cdots d_{N-1,N}\end{array}\right),{\hat{u}}_{2,N}\left(\begin{array}{cc}{d}_{1,N}& d_{2,N}\cdots d_{N-1,N}\end{array}\right),\cdots ,{\hat{u}}_{M,N}\left(\begin{array}{cc}{d}_{1,N}& d_{2,N}\cdots d_{N-1,N}\end{array}\right)\right)}^{\text{T}}.\end{array}$

这里“⊗”表示Kronecker积， $E_n$ 表示n阶单位矩阵，则(10)式可化为等价的非线性方程组形式：

$\left(A\otimes E_{N-1}-E_M\otimes C\right)Y=-Y.\ast \left(\left(E_M\otimes B\right)Y\right)-R-Y.*H_2+H_1.$

则不动点迭代格式为：

$Y_{n+1}=-{\left(A\otimes E_{N-1}-E_M\otimes C\right)}^{-1}\left[Y_n.\ast \left(\left(E_M\otimes B\right)Y_n\right)+R+Y.*H_2-H_1\right],n=0,1,2,\cdots .$ (11)

用格式(11)求解(6)式。在(1)式中令 $a=0,b=5,T=10$。用 $L^{\infty }$ -来度量数值误差：

$E_{M,N}=\underset{1\le l\le M,1\le k\le N-1}{\max }\left|u\left(x_k,t_l\right)-u_{M,N}\left(x_k-t_l\right)\right|.$

由文献 [12] Burgers方程的精确解为：

$\begin{array}{l}u\left(x,t\right)=\frac{2\frac{\omega }{k}{\text{e}}^{\frac{-\omega }{\alpha k^2}\left(kx-\omega t\right)}}{C_1+{\text{e}}^{\frac{-\omega }{\alpha k^2}\left(kx-\omega t\right)}},\phi \left(x\right)=\frac{2\frac{\omega }{k}{\text{e}}^{\frac{-\omega }{\alpha k^2}kx}}{C_1+{\text{e}}^{\frac{-\omega }{\alpha k^2}kx}},\\ g_1\left(t\right)=\frac{2\frac{\omega }{k}{\text{e}}^{\frac{-\omega }{\alpha k^2}\left(ak-\omega t\right)}}{C_1+{\text{e}}^{\frac{-\omega }{\alpha k^2}\left(ak-\omega t\right)}},g_2\left(t\right)=\frac{2\frac{\omega }{k}{\text{e}}^{\frac{-\omega }{\alpha k^2}\left(bk-\omega t\right)}}{C_1+{\text{e}}^{\frac{-\omega }{\alpha k^2}\left(bk-\omega t\right)}}.\end{array}$ (12)

这里 $k,\alpha ,\omega ,C_1$，都是常数。

图1是令(12)式中的参数 $\alpha =1,\omega =0.2,k=0.45,C_1=\left(2\ast w\right)/k,M=14$ 时，最大误差 $E_{M,N}$ 随空间方向插值次数N的变化情况。可以看出，随着N的增大，误差递减，说明所提算法格式在空间方向逼近效果良好，有效证明了所提算法的优越性。

图2中是不改变图一参数时，固定时间方向多项式次数 $N=20$，最大误差 $E_{M,N}$ 随时间插值多项式次数M的变化情况。表明本文算法在时间方向逼近Burgers方程精确解也具有有效性与高精度。

Figure 1. $L^{\infty }$ -error with $M=14,\alpha =1,w=0.2,k=0.45,C_1=\left(2\ast w\right)/k$

图1. $M=14,\alpha =1,w=0.2,k=0.45,C_1=\left(2\ast w\right)/k$ 时的 $L^{\infty }$ -误差

Figure 2. $L^{\infty }$ -error with $N=20,\alpha =1,w=0.2,k=0.45,C_1=\left(2\ast w\right)/k$

图2. $N=20,\alpha =1,w=0.2,k=0.45,C_1=\left(2\ast w\right)/k$ 时的 $L^{\infty }$ -误差

4. 结论

针对Burgers方程的混合问题，我们采用等距节点的Lagrange插值逼近方法构造了时空方向的Lagrange插值逼近格式，对Burgers方程一步一步推导转化为矩阵方程，采用不动点迭代方法进行求解。数值实验结果表明近似解能较好地吻合方程的精确解，而且算法格式简单有效，所使用的时间和空间的节点数相差不大，能够极大提高计算工作效率。另外，由于算法格式简单，易于理解，数值实验结果良好，所以本文所提算法对解决科学工程中其他问题也有着广泛应用性。

基金项目

国家自然科学基金(No. 11371123，No. 11771299)，河南省自然科学基金(No. 202300410156)。