指标1-型积分代数方程的配置边值方法

A Collocation Boundary Value Method for Integral Algebraic Equations of Index-1

摘要: 通过构造配置边值方法求解指标-1型积分代数方程。利用拉格朗日插值公式，选取未计算的节点作插值节点，将原方程离散为一个线性方程组。由绕圈数以及剩余理论分析了该方法的可解性和收敛性。证明了该方法求解指标-1型积分代数方程具有较高的收敛阶，同时给出数值算例验证了我们的理论结果。

Abstract: The index-1 integral algebraic equation is solved by constructing boundary value method. Using Lagrange interpolation formula, the nodes that not calculated are selected as interpolation nodes, then original equation is discreted into a system of linear equations. And the solvability and con-vergence of the method are analyzed based on the winding number and residual theory. It is proved that this method has higher convergence order for solving index-1 type integral algebraic equations, and numerical examples are given to verify our theoretical results.

1. 引言

由第一类与第二类Volterra积分方程组成的积分代数方程在各个自然科学学科和数学建模中有着大量的应用 [1] [2] [3] [4]。而积分代数方程是一个庞大的系统，对该方程的求解是比较复杂的。对此，研究者引入指标来协助求解积分代数方程。Gear引入了积分代数方程的“指标”的概念 [5]，该说法类似于用数据来评估积分代数方程 [6]。在此之后，很多学者提出了大量求解积分代数方程的数值方法 [7] [8] [9] [10]，但均未对指标进行深入的研究。最近Liang和Brunner通过引入线性积分代数方程的指标填补了这一空缺 [11]，给出了积分代数方程的指标的定义以及如何求得其指标的计算方法，提出指标-1型积分代数方程可以解耦成如下形式

$\{\begin{array}{l}u\left(t\right)+{\displaystyle {\int }_0^t{K}_{11}\left(t,s\right)u\left(s\right)\text{d}s}+{\displaystyle {\int }_0^t{K}_{12}\left(t,s\right)v\left(s\right)\text{d}s=f\left(t\right),}\\ {\displaystyle {\int }_0^t{K}_{21}\left(t,s\right)u\left(s\right)\text{d}s+{\displaystyle {\int }_0^t{K}_{22}\left(t,s\right)v\left(s\right)\text{d}s=g\left(t\right).}}\end{array}$ (1)

其中， $t\in \left[0,T\right]$， $\left|K_{22}\left(t,t\right)\right|\ne 0$，函数 $f,g$ 以及核函数 $K_{i,j}\left(t,s\right),\left(i,j=1,2\right)$ 足够光滑， $g\left(0\right)=0$。并证明了方程(1)有唯一解，给出该方程一般系统的配置解。

随后，Liang和Brunner分析了指标-2型积分代数方程的解耦系统，证明了配置法在给定的系统和解耦后的系统中的应用是一致的 [12]。基于对指标- $\mu$ 型积分代数方程解耦成新的系统，其理论分析以及数值计算方法已经有了大量的结果 [13] [14] [15] [16]。配置边值方法是求解常微分方程的一种方法 [17]，Chen和Zhang利用该方法来求解Volterra积分方程 [18] [19]。Ma和Xiang基于特殊的多步配置方法，利用未计算的近似值，研究了求解Volterra积分方程的k步配置边值方法(CBVM-k) [20]。Liu和Ma进一步利用该理论构造了第一类Volterra积分方程的块配置边值方法 [21]。他们都得到了具有高阶收敛的数值解，并且具有很强的稳定性。这一思想对本论文的研究有着重要的影响。

在本文中，我们考虑利用CBVM-k求解指标-1型积分代数方程

$\{\begin{array}{l}u\left(t\right)+{\displaystyle {\int }_0^t{K}_{11}\left(t-s\right)u\left(s\right)\text{d}s+{\displaystyle {\int }_0^t{K}_{12}\left(t-s\right)v\left(s\right)\text{d}s=f\left(t\right),}}\\ {\displaystyle {\int }_0^t{K}_{21}\left(t-s\right)u\left(s\right)\text{d}s+{\displaystyle {\int }_0^t{K}_{22}\left(t-s\right)v\left(s\right)\text{d}s=g\left(t\right).}}\end{array}$ (2)

其中 $K_{22}\left(0\right)\ne 0$。并讨论CBVM-k求解指标-1型积分代数方程的解的在唯一性以及收敛性。

本文工作安排如下，在第二节，我们构造指标-1型积分代数方程的k步配置边值方法(CBVM-k)。第三节，我们利用Toeplitz矩阵的一些理论进行可解性和收敛性分析。在第四节给出数值例子验证我们的理论分析。最后，我们进行简要的总结。

2. k步配置边值方法

在这一节中，我们将构造指标-1型积分代数方程(1)的k步配置边值方法(CBVM-k)。

对区间 $\left[0,T\right]$ 作均匀网格划分

$X_h=\left\{t_n:t_n=nh,n=0,1,\cdots ,N,h=\frac{T}{N}\right\}$

在区间 $\left[t_n,t_{n+1}\right]$ 上 $\left(n<N-k\right)$，选取 $t_n,t_{n+1}$ 以及 $t_{n+1}$ 的后k个节点为配置节点，在 $\left[t_{N-k},t_N\right]$ 用 $t_{N-k},\cdots ,t_N$ 为配置节点。定义差值基函数

${\varphi }_j^k\left(s\right)={\displaystyle \underset{i=0,i\ne j}{\overset{k+1}{\prod }}\frac{s-i}{j-i}},$

则，当 $n=1,2,\cdots ,N-k-1$ 时，我们有

$f\left(t_n\right)=u_h\left(t_n\right)+h{\displaystyle \underset{j=0}{\overset{n-1}{\sum }}{\displaystyle {\int }_0^1{K}_{11}\left(t_n-\left(t_j+sh\right)\right){\displaystyle \underset{i=0}{\overset{k+1}{\sum }}{\varphi }_i^k}\left(s\right)\text{d}s{U}_{j+i}}}+h{\displaystyle \underset{j=0}{\overset{n-1}{\sum }}{\displaystyle {\int }_0^1{K}_{12}\left(t_n-\left(t_j+sh\right)\right){\displaystyle \underset{i=0}{\overset{k+1}{\sum }}{\varphi }_i^k}\left(s\right)\text{d}s{V}_{j+i}}},$

$g\left(t_n\right)=h{\displaystyle \underset{j=0}{\overset{n-1}{\sum }}{\displaystyle {\int }_0^1{K}_{21}\left(t_n-\left(t_j+sh\right)\right){\displaystyle \underset{i=0}{\overset{k+1}{\sum }}{\varphi }_i^k}\left(s\right)\text{d}s{U}_{j+i}}}+h{\displaystyle \underset{j=0}{\overset{n-1}{\sum }}{\displaystyle {\int }_0^1{K}_{22}\left(t_n-\left(t_j+sh\right)\right){\displaystyle \underset{i=0}{\overset{k+1}{\sum }}{\varphi }_i^k}\left(s\right)\text{d}s{V}_{j+i}}},$

当 $n=N-k,\cdots ,N$ 时，我们有

$\begin{array}{c}f\left(t_n\right)=U_n+h{\displaystyle \underset{j=0}{\overset{N-k-2}{\sum }}{\displaystyle {\int }_0^1{K}_{11}\left(t_n-\left(t_j+sh\right)\right){\displaystyle \underset{i=0}{\overset{k+1}{\sum }}{\varphi }_i^k}\left(s\right)U_{j+i}\text{d}s}}\\ +h{\displaystyle \underset{j=0}{\overset{N-k-2}{\sum }}{\displaystyle {\int }_0^1{K}_{12}\left(t_n-\left(t_j+sh\right)\right){\displaystyle \underset{i=0}{\overset{k+1}{\sum }}{\varphi }_i^k}\left(s\right)V_{j+i}\text{d}s}}\\ +h{\displaystyle {\int }_0^{n-N+k+1}{K}_{11}\left(t_n-\left(t_{N-k-1}+sh\right)\right){\displaystyle \underset{i=0}{\overset{k}{\sum }}{\varphi }_i^{k-1}}\left(s\right)U_{N-k+i}\text{d}s}\\ +h{\displaystyle {\int }_0^{n-N+k+1}{K}_{12}\left(t_n-\left(t_{N-k-1}+sh\right)\right){\displaystyle \underset{i=0}{\overset{k}{\sum }}{\varphi }_i^{k-1}}\left(s\right)V_{N-k+i}\text{d}s},\end{array}$

$\begin{array}{c}g\left(t_n\right)=h{\displaystyle \underset{j=0}{\overset{N-k-2}{\sum }}{\displaystyle {\int }_0^1{K}_{21}\left(t_n-\left(t_j+sh\right)\right){\displaystyle \underset{i=0}{\overset{k+1}{\sum }}{\varphi }_i^k}\left(s\right)U_{j+i}\text{d}s}}\\ +h{\displaystyle \underset{j=0}{\overset{N-k-2}{\sum }}{\displaystyle {\int }_0^1{K}_{22}\left(t_n-\left(t_j+sh\right)\right){\displaystyle \underset{i=0}{\overset{k+1}{\sum }}{\varphi }_i^k}\left(s\right)V_{j+i}\text{d}s}}\\ +h{\displaystyle {\int }_0^{n-N+k+1}{K}_{21}\left(t_n-\left(t_{N-k-1}+sh\right)\right){\displaystyle \underset{i=0}{\overset{k}{\sum }}{\varphi }_i^{k-1}}\left(s\right)U_{N-k+i}\text{d}s}\\ +h{\displaystyle {\int }_0^{n-N+k+1}{K}_{22}\left(t_n-\left(t_{N-k-1}+sh\right)\right){\displaystyle \underset{i=0}{\overset{k}{\sum }}{\varphi }_i^{k-1}}\left(s\right)V_{N-k+i}\text{d}s}.\end{array}$

在介绍CBVM-k之前，我们先引入一些记号，令 ${\alpha }_m\left(t\right)={\displaystyle \underset{i=-m+1}{\overset{m-1}{\sum }}{a}_i{t}^i}$，我们定义一个m阶的Toeplitz矩阵如下

$T\left[{\alpha }_m\right]=\left[\begin{array}{cccc}{a}_0& a_{-1}& \cdots & a_{-m+1}\\ a_{-1}& a_{-0}& \cdots & a_{-m+2}\\ ⋮& ⋮& \ddots & ⋮\\ a_{m-1}& a_{m-2}& \cdots & a_0\end{array}\right].$

定义

$b_{-j,i}^{pq}={\displaystyle {\int }_0^1{K}_{pq}\left(h\left(j-s\right)\right){\varphi }_i^k\left(s\right)\text{d}s},{\hat{b}}_{j,i}^{pq}={\displaystyle {\int }_0^{j-N+k+1}{K}_{pq}\left(h\left(j-N+k+1\right)\right){\varphi }_i^{k-1}\left(s\right)\text{d}s,}$

其中 $p,q=1,2$。定义Laurent多项式如下

${\beta }_m^{pq,i}\left(t\right)={\displaystyle \underset{j=-m+1}{\overset{0}{\sum }}{b}_{j-1,i}^{pq}{t}^j},i=0,1,\cdots ,k+1.$

我们有Toeplitz型矩阵

$H_i^{pq}=\left(0_{N\times i}T\left[{\beta }_N^{pq,i}\left(t\right)\right]\left(1:N,1:N-k-1\right)0_{N\times \left(k-i+2\right)}\right),i=0,1,2,\cdots ,k+1.$

以及一个 $N\times \left(N+1\right)$ 的矩阵 $S^{pq}$，并且

$\begin{array}{l}{S}^{pq}\left(N-k:N,N-k+1:N+1\right)\\ =\left[\begin{array}{cccc}{\hat{b}}_{N-k,0}^{pq}& {\hat{b}}_{N-k,1}^{pq}& \cdots & {\hat{b}}_{N-k,k}^{pq}\\ {\hat{b}}_{N-k+1,0}^{pq}& {\hat{b}}_{N-k+1,1}^{pq}& \cdots & {\hat{b}}_{N-k+1,k}^{pq}\\ ⋮& ⋮& \ddots & ⋮\\ {\hat{b}}_{N,0}^{pq}& {\hat{b}}_{N,1}^{pq}& \cdots & {\hat{b}}_{N,k}^{pq}\end{array}\right]\end{array}$

其他位置的元素为0。令

${\bar{A}}^{pq}=S^{pq}+{\displaystyle \underset{i=0}{\overset{k+1}{\sum }}{H}_i^{pq}},A_{pq}={\bar{A}}^{pq}\left(1:N,2:N+1\right),{\bar{a}}^{pq}=h{\bar{A}}^{pq}\left(1:N,1:1\right),$

$F={\left(f\left(t_1\right),f\left(t_2\right),\cdots ,f\left(t_N\right)\right)}^{\text{T}},G={\left(g\left(t_1\right),g\left(t_2\right),\cdots ,g\left(t_N\right)\right)}^{\text{T}},$

$U=\left(U_1,U_2,\cdots ,U_N\right),V=\left(V_1,V_2,\cdots ,V_N\right).$

我们可以得到

$\left[\begin{array}{cc}I+h{A}_{11}& h{A}_{12}\\ h{A}_{21}& h{A}_{22}\end{array}\right]\left[\begin{array}{c}U\\ V\end{array}\right]=\left[\begin{array}{c}F\\ G\end{array}\right]-\left[\begin{array}{c}{U}_0{\bar{a}}_{11}+V_0{\bar{a}}_{12}\\ U_0{\bar{a}}_{21}+V_0{\bar{a}}_{22}\end{array}\right].$ (3)

3. 可解性与收敛性分析

在这一节，我们讨论配置边值方法求解指标-1型积分代数方程解的存在唯一性与收敛性。传统配置法通过迭代求得近似解，而配置边值方法的近似解是利用未计算的近似值，通过一个线性系统整体求解而来，因此我们需要对该方法的存在唯一性、收敛性进行新的讨论。

首先，我们讨论积分代数方程(2)中核 $K_{22}\left(t\right)=1$ 这一特殊情况。配置边值方法求解这一特殊情况解的存在唯一性、收敛性理论有助于我们讨论配置边值方法解一般情况下方程(2)的解的存在唯一性、收敛性。

3.1. $K_{22}\left(t\right)=1$

当 $K_{22}\left(t\right)=1$，我们有

$\begin{array}{c}f\left(t_n\right)=u_h\left(t_n\right)+h{\displaystyle \underset{j=0}{\overset{n-1}{\sum }}{\displaystyle {\int }_0^1{K}_{11}\left(t_n-\left(t_j+sh\right)\right)u_h\left(t_j+sh\right)\text{d}s}}\\ +h{\displaystyle \underset{j=0}{\overset{n-1}{\sum }}{\displaystyle {\int }_0^1{K}_{12}\left(t_n-\left(t_j+sh\right)\right)v_h\left(t_j+sh\right)\text{d}s}},\end{array}$ (4)

$g\left(t_n\right)=h{\displaystyle \underset{j=0}{\overset{n-1}{\sum }}{\displaystyle {\int }_0^1{K}_{21}\left(t_n-\left(t_j+sh\right)\right)u_h\left(t_j+sh\right)\text{d}s}}+h{\displaystyle \underset{j=0}{\overset{n-1}{\sum }}{\displaystyle {\int }_0^1{v}_h\left(t_j+sh\right)\text{d}s}},$

当 $n=N-1,N-1,\cdots ,0$ 时， $g\left(t_{n+1}\right)-g\left(t_n\right)$ 有

$\begin{array}{c}g\left(t_{n+1}\right)-g\left(t_n\right)=h{\displaystyle \underset{j=0}{\overset{n}{\sum }}{\displaystyle {\int }_0^1{K}_{21}\left(t_{n+1}-\left(t_j+sh\right)\right)u_h\left(t_j+sh\right)\text{d}s}}\\ -h{\displaystyle \underset{j=0}{\overset{n-1}{\sum }}{\displaystyle {\int }_0^1{K}_{21}\left(t_n-\left(t_j+sh\right)\right)u_h\left(t_j+sh\right)\text{d}s}}+h{\displaystyle {\int }_0^1{v}_h\left(t_n+sh\right)\text{d}s},\end{array}$ (5)

则方程(3)可以改写为

$\left[\begin{array}{cc}I+h{A}_{11}& h{A}_{12}\\ h{\hat{A}}_{21}& h{B}_N^k\end{array}\right]\left[\begin{array}{c}U\\ V\end{array}\right]=\left[\begin{array}{c}F\\ \hat{G}\end{array}\right]-\left[\begin{array}{c}{U}_0{\bar{a}}_{11}+V_0{\bar{a}}_{12}\\ U_0{\hat{a}}_{21}+V_0{\hat{a}}_{22}\end{array}\right],$ (6)

当 $h\to 0$ 是对系数矩阵高斯消元有

$\left[\begin{array}{cc}I+h{A}_{11}& h{A}_{12}\\ h{\hat{A}}_{21}& h{B}_N^k\end{array}\right]\to \left[\begin{array}{cc}I+O\left(h\right)& O\left(h\right)\\ 0& h{B}_N^k\end{array}\right].$ (7)

其中 $\hat{G}={\left[g\left(t_1\right)-g\left(t_0\right),\cdots ,g\left(t_N\right)-g\left(t_{N-1}\right)\right]}^{\text{T}},$

${\hat{a}}^{21}={\bar{A}}^{21}\left(1:N,1:1\right)-{\left[\text{0}{\left({\bar{A}}^{21}\left(1:N-1,1:1\right)\right)}^{\text{T}}\right]}^{\text{T}},$

${\hat{a}}^{22}={\left[h{\displaystyle {\int }_0^1{\varphi }_0^k\left(s\right)\text{d}s,0,\cdots ,0}\right]}_{1\times N}^{\text{T}},{\hat{A}}_{21}=A_{21}-{\left(0,A_{21}{\left(1:N,:\right)}^{\text{T}}\right)}^{\text{T}}.$

$B_N^k=\left[\begin{array}{cc}{T}_N^k& r\\ 0& R_N^k\end{array}\right]$，其中 $r$ 是一个 $\left(N-k-1\right)\times \left(k+1\right)$ 矩阵， $T_N^k$ 是一个由Laurent多项式 ${\varsigma }_{N-k-1}^{22}\left(t\right)={\displaystyle \underset{j=-k}{\overset{1}{\sum }}{b}_{1,1-i}^{22}{t}^j}$ 定义的 $\left(N-k-1\right)\times \left(N-k-1\right)$ Toeplitz矩阵，

$R_N^k=\left[\begin{array}{cccc}{\hat{b}}_{N-k,0}^{21}& {\hat{b}}_{N-k,1}^{21}& \cdots & {\hat{b}}_{N-k,k}^{21}\\ {\hat{b}}_{N-k+1,0}^{21}& {\hat{b}}_{N-k+1,1}^{21}& \cdots & {\hat{b}}_{N-k+1,k}^{21}\\ ⋮& ⋮& \ddots & ⋮\\ {\hat{b}}_{N,0}^{21}& {\hat{b}}_{N,1}^{21}& \cdots & {\hat{b}}_{N,k}^{21}\end{array}\right].$

因此方程(4)有唯一解的充要条件是 $T_N^k$ 和 $R_N^k$ 可逆。现在我们讨论 $T_N^k$ 与 $R_N^k$ 的可逆性，假设Laurent多项式 $d\left(t\right)={\displaystyle \underset{n=-\infty }{\overset{\infty }{\sum }}{d}_n{t}^n}$ 所定义的无穷维Toeplitz矩阵如下：

$T\left[d\right]=\left[\begin{array}{cccc}{b}_0& b_{-1}& b_{-2}& \cdots \\ b_1& b_0& b_{-1}& \cdots \\ b_2& b_1& b_0& \cdots \\ ⋮& ⋮& ⋮& \ddots \end{array}\right].$

令 $T$ 是复平面上的单位圆盘，t在 $T$ 逆时针绕原点一圈时， $d\left(t\right)$ 绘制一条连续的封闭曲线。这条曲线绕原点逆时针旋转的次数表示 $d\left(t\right)$ 的圈数，记作 $Win{d}_d$。特别地，假设对于所有的 $t\in T$， $d\left(t\right)\ne 0$，且 $d\left(t\right)$ 的非零系数是可数的，则我们有，

$d\left(t\right)={\displaystyle \underset{j=-r}{\overset{s}{\sum }}{d}_j{t}^j},$

进一步我们可以直接得到如下形式：

$d\left(t\right)=t^{-r}{b}_s{\displaystyle \underset{j=1}{\overset{J}{\prod }}\left(t-{\delta }_j\right)}{\displaystyle \underset{i=1}{\overset{I}{\prod }}\left(t-{\mu }_i\right)}=t^{-r}\bar{d}\left(t\right),$

又因为所有的 $t\in T$， $d\left(t\right)\ne 0$，所以 $T\left[d\right]$ 是一个可逆紧模算子，则存在一个算子D，使得 $DT\left[d\right]-I$ 和 $T\left[d\right]D-I$ 是紧的 [22]。基于此我们有如下结果。

引理1 算子 $T\left[d\right]$ 是可逆的当且仅当对于所有的 $t\in T$， $d\left(t\right)\ne 0$，以及 $Win{d}_d=0$。

记 $T_n\left[d\right]$ 表示 $T\left[d\right]\left(1:n,1:n\right)$，我们有 $\underset{n\to \infty }{\lim }{T}_n\left[d\right]=T\left[d\right]$， $T_n\left[d\right]$ 的可逆性由 $T\left[d\right]$ 决定。

引理2 设 $d\left(t\right)$ 属于Wiener代数。则，

当 $T\left[d\right]$ 可逆时， $\underset{n\to \infty }{\lim }\sup \Vert {\left(T_n\left[d\right]\right)}^{-1}\Vert <\infty$，

当 $T\left[d\right]$ 不可逆时， $\underset{n\to \infty }{\lim }\sup \Vert {\left(T_n\left[d\right]\right)}^{-1}\Vert =\infty$。

另一方面，由于局部插值函数是线性无关的，所以 $R_N^k$ 是可逆的。因此，当 ${\varsigma }_{N-k-1}^{22}\left(t\right)$ 属于Wiener代数且 $Win{d}_{{\varsigma }_{N-k-1}^{22}\left(t\right)}=0$ 时，方程(6)的系数矩阵是可逆的。

由绕圈数以及剩余理论，我们将在下面的定理中给出配置边值方法求解指标-1型积分代数方程解存在的条件和收敛阶。

定理1 设方程(2)是指标-1型积分代数方程，假设 $f\in C^{k+1}\left(\left[0,T\right]\right)$， $g\in C^{k+2}\left(\left[0,T\right]\right)$ 且 $g\left(0\right)=0$， $K_{21}\in C^{k+2}(\left(\left[0,T\right]\right)$， $K_{22}\left(t\right)=1$， $K_{1q}\in C^{k+1}\left(\left[0,T\right]\right)$， $\left(q=1,2\right)$，并且假设 ${\varsigma }_{N-k-1}^{22}\left(t\right)$ 的绕圈数为0，则CBVM-k有唯一解，配置误差在配置节点有界且满足

${\Vert u-u_h\Vert }_{\infty }=O\left(h^{k+2}\right),$

${\Vert v-v_h\Vert }_{\infty }=O\left(h^{k+1}\right)$

证明 因为 ${\varsigma }_{N-k-1}^{22}\left(t\right)$ 的绕圈数为0，有引理1和引理2可知(7)式中 $B_N^k$ 是可逆的，因此方程(6)有唯一解，即方程(3)有唯一解。

下面我们考虑CBVM-k的收敛性。令 $e_h=u\left(t\right)-u_h\left(t\right)$， ${\tilde{e}}_h=v\left(t\right)-v_h\left(t\right)$，我们可以的到误差方程

$0=e_h\left(t_n\right)+h{\displaystyle \underset{j=0}{\overset{n-1}{\sum }}{\displaystyle {\int }_0^1{K}_{11}\left(t_n-\left(t_j+sh\right)\right)e_h\left(t_j+sh\right)\text{d}s}}+h{\displaystyle \underset{j=0}{\overset{n-1}{\sum }}{\displaystyle {\int }_0^1{K}_{12}\left(t_n-\left(t_j+sh\right)\right){\tilde{e}}_h\left(t_j+sh\right)\text{d}s}},$

$0=h{\displaystyle \underset{j=0}{\overset{n-1}{\sum }}{\displaystyle {\int }_0^1{K}_{21}\left(t_n-\left(t_j+sh\right)\right)e_h\left(t_j+sh\right)\text{d}s{U}_{j+i}}}+h{\displaystyle \underset{j=0}{\overset{n-1}{\sum }}{\displaystyle {\int }_0^1{\tilde{e}}_h\left(t_j+sh\right)\text{d}s}},$

利用对(5)同样的处理方式有

$\begin{array}{c}0=h{\displaystyle \underset{j=0}{\overset{n}{\sum }}{\displaystyle {\int }_0^1{K}_{21}\left(t_{n+1}-\left(t_j+sh\right)\right)e_h\left(t_j+sh\right)\text{d}s}}\\ -h{\displaystyle \underset{j=0}{\overset{n-1}{\sum }}{\displaystyle {\int }_0^1{K}_{21}\left(t_n-\left(t_j+sh\right)\right)e_h\left(t_j+sh\right)\text{d}s}}+h{\displaystyle \underset{j=0}{\overset{n-1}{\sum }}{\displaystyle {\int }_0^1{\tilde{e}}_h\left(t_j+sh\right)\text{d}s}},\end{array}$

由Peano核定理有，当 $j=0,1,2,\cdots ,N-k-2$，在 $t_j,t_{j+1},\cdots ,t_{j+k+1}$ 处有

$e_h\left(t_j+sh\right)={\displaystyle \underset{i=0}{\overset{k+1}{\sum }}{\varphi }_i^k\left(s\right)e_h\left(t_{j+i}\right)}+O\left(h^{k+2}\right),{\tilde{e}}_h\left(t_j+sh\right)={\displaystyle \underset{i=0}{\overset{k+1}{\sum }}{\varphi }_i^k\left(s\right){\tilde{e}}_h\left(t_{j+i}\right)}+O\left(h^{k+2}\right)$

当 $j=N-k-1,\cdots ,N-1$ 时有

$e_h\left(t_j+sh\right)={\displaystyle \underset{i=0}{\overset{k}{\sum }}{\varphi }_i^{k-1}\left(s\right)e_h\left(t_{N-k+i}\right)}+O\left(h^{k+1}\right),{\tilde{e}}_h\left(t_j+sh\right)={\displaystyle \underset{i=0}{\overset{k}{\sum }}{\varphi }_i^{k-1}\left(s\right){\tilde{e}}_h\left(t_{N-k+i}\right)}+O\left(h^{k+1}\right)$

记 $E_i=e_h\left(t_i\right)$， ${\tilde{E}}_i={\tilde{e}}_h\left(t_i\right)$， $E={\left[E_1,\cdots ,E_N\right]}^{\text{T}}$， $\tilde{E}={\left[{\tilde{E}}_1,\cdots ,{\tilde{E}}_N\right]}^{\text{T}}$

有

$\left[\begin{array}{cc}I+h{A}_{11}& h{A}_{12}\\ h{\hat{A}}_{21}& h{B}_N^k\end{array}\right]\left[\begin{array}{c}E\\ \tilde{E}\end{array}\right]=\left[\begin{array}{l}{r}_1\\ r_2\end{array}\right],$

其中 $r_1=r_2={\left[O\left(h^{k+2}\right),\cdots ,O\left(h^{k+2}\right)\right]}_{1\times N}^{\text{T}}$， $B_N^k$ 与(6)中的一致。则由(7)有

$\left[\begin{array}{cc}I+O\left(h\right)& h{A}_{12}\\ 0& B_N^k\end{array}\right]\left[\begin{array}{c}E\\ \tilde{E}\end{array}\right]=\left[\begin{array}{c}{r}_1\\ {\tilde{r}}_2+r_1\end{array}\right]$

${\tilde{r}}_2={\left[O\left(h^{k+1}\right),\cdots ,O\left(h^{k+1}\right)\right]}_{1\times N}^{\text{T}}$，由引理2有，当 $N\to \infty$ 时 $B_N^k$ 是有界的，所以

${\Vert u-u_h\Vert }_{\infty }=O\left(h^{k+2}\right),$

${\Vert v-v_h\Vert }_{\infty }=O\left(h^{k+1}\right).$

证毕。

3.2. 一般情况下的核函数 $K_{22}(\; t\; )$

在这一小节，我们考虑一般情况下的核函数 $K_{22}\left(t\right)$，且核函数可以表示为

$K_{22}\left(t\right)=K_{22}\left(0\right)+t{\bar{K}}_{22}\left(t\right),$

并且 $K_{22}\left(0\right)\ne 0$，我们有以下的理论。

定理2 设方程(2)是指标-1型积分代数方程，假设 $f\in C^{k+1}\left(\left[0,T\right]\right)$， $g\in C^{k+2}\left(\left[0,T\right]\right)$ 且 $g\left(0\right)=0$， $K_{1q}\in C^{k+1}\left(\left[0,T\right]\right)$， $K_{2q}\in C^{k+2}\left(\left[0,T\right]\right)$， $\left(q=1,2\right)$， $K_{22}\left(0\right)\ne 0$，并且假设 ${\varsigma }_{N-k-1}^{22}\left(t\right)$ 的绕圈数为0。则CBVM-k有唯一解，配置误差在配置节点有界且满足

${\Vert u-u_h\Vert }_{\infty }=O\left(h^{k+2}\right),$

${\Vert v-v_h\Vert }_{\infty }=O\left(h^{k+1}\right).$

证明 当 $n=1,2,\cdots ,N$ 时，我们有

$\begin{array}{c}f\left(t_n\right)=u_h\left(t_n\right)+h{\displaystyle \underset{j=0}{\overset{n-1}{\sum }}{\displaystyle {\int }_0^1{K}_{11}\left(t_n-\left(t_j+sh\right)\right)u_h\left(t_j+sh\right)\text{d}s}}\\ +h{\displaystyle \underset{j=0}{\overset{n-1}{\sum }}{\displaystyle {\int }_0^1{K}_{12}\left(t_n-\left(t_j+sh\right)\right)v_h\left(t_j+sh\right)\text{d}s}},\end{array}$ (8)

$g\left(t_n\right)=h{\displaystyle \underset{j=0}{\overset{n-1}{\sum }}{\displaystyle {\int }_0^1{K}_{21}\left(t_n-\left(t_j+sh\right)\right)u_h\left(t_j+sh\right)\text{d}s}}+h{\displaystyle \underset{j=0}{\overset{n-1}{\sum }}{\displaystyle {\int }_0^1{K}_{22}\left(t_n-\left(t_j+sh\right)\right)v_h\left(t_j+sh\right)\text{d}s}},$

当 $n=N-1,N-1,\cdots ,0$ 时，由 $g\left(t_{n+1}\right)-g\left(t_n\right)$ 有

$\begin{array}{l}g\left(t_{n+1}\right)-g\left(t_n\right)\\ =h{\displaystyle \underset{j=0}{\overset{n}{\sum }}{\displaystyle {\int }_0^1{K}_{21}\left(t_{n+1}-\left(t_j+sh\right)\right)u_h\left(t_j+sh\right)\text{d}s}}-h{\displaystyle \underset{j=0}{\overset{n-1}{\sum }}{\displaystyle {\int }_0^1{K}_{21}\left(t_n-\left(t_j+sh\right)\right)u_h\left(t_j+sh\right)\text{d}s}}\\ +h{\displaystyle \underset{j=0}{\overset{n}{\sum }}{\displaystyle {\int }_0^1{K}_{22}\left(t_{n+1}-\left(t_j+sh\right)\right)v_h\left(t_j+sh\right)\text{d}s}}-h{\displaystyle \underset{j=0}{\overset{n-1}{\sum }}{\displaystyle {\int }_0^1{K}_{22}\left(t_n-\left(t_j+sh\right)\right)v_h\left(t_j+sh\right)\text{d}s}},\end{array}$ (9)

代入插值多项式，由(8)，(9)有

$\left[\begin{array}{cc}I+h{A}_{11}& h{A}_{12}\\ h{\hat{A}}_{21}& h{\hat{A}}_{22}\end{array}\right]\left[\begin{array}{c}U\\ V\end{array}\right]=\left[\begin{array}{c}F\\ \hat{G}\end{array}\right]-\left[\begin{array}{c}{U}_0{\bar{a}}_{11}+V_0{\bar{a}}_{12}\\ U_0{\hat{a}}_{21}+V_0{\hat{a}}_{22}\end{array}\right],$ (10)

其中

$\hat{G}={\left[g\left(t_1\right)-g\left(t_0\right),\cdots ,g\left(t_N\right)-g\left(t_{N-1}\right)\right]}^{\text{T}},$

${\hat{A}}_{21}$ 的构造与(6)式中一致。则由高斯消元法有，

$\left[\begin{array}{cc}I+h{A}_{11}& h{A}_{12}\\ h{\hat{A}}_{21}& h{\hat{A}}_{22}\end{array}\right]\to \left[\begin{array}{cc}I+O\left(h\right)& h{A}_{12}\\ 0& h{\hat{A}}_{22}\end{array}\right].$ (11)

现在我们讨论 ${\hat{A}}_{22}$ 的奇异性，我们知道 ${\hat{A}}_{22}$ 的构造方式与 ${\hat{A}}_{21}$ 一致。

记 $W_{i,j}^1={\displaystyle {\int }_0^1{K}_{21}\left(t_i-\left(t_j+sh\right)\right)u_h\left(t_j+sh\right)\text{d}s}$， $W_{i,j}^2={\displaystyle {\int }_0^1{K}_{22}\left(t_i-\left(t_j+sh\right)\right)v_h\left(t_j+sh\right)\text{d}s}$，

由(9)式有

$\begin{array}{l}h\left[\begin{array}{cccccccc}{W}_{1,0}^1& 0& \cdots & 0& W_{1,0}^2& 0& \cdots & 0\\ W_{2,0}^1-W_{1,0}^1& W_{2,1}^1& \cdots & 0& W_{2,0}^2-W_{1,0}^2& W_{2,1}^2& \cdots & 0\\ ⋮& ⋮& \cdots & ⋮& ⋮& ⋮& \ddots & ⋮\\ W_{N,0}^1-W_{N-1,0}^1& W_{N,1}^1-W_{N-1,1}^1& \cdots & W_{N,N-1}^1& W_{N,0}^2-W_{N-1,0}^2& W_{N,1}^2-W_{N-1,1}^2& \cdots & W_{N,N-1}^2\end{array}\right]\left[\begin{array}{l}1\\ 1\end{array}\right]\\ =\left[\begin{array}{c}g\left(t_1\right)\\ g\left(t_2\right)-g\left(t_1\right)\\ ⋮\\ g\left(t_N\right)-g\left(t_{N-1}\right)\end{array}\right],\end{array}$

当 $h\to 0$ 时， $W_{i,j}^1\to O\left(1\right)$，由中值定理有， $W_{i,j-1}^2-W_{i-1,j-1}^2\to O\left(h\right)$，则由高斯消元法有

$h\left[\begin{array}{cc}{W}^1& W^2\end{array}\right]\left[\begin{array}{l}1\\ 1\end{array}\right]=\left[\tilde{G}\right],$

其中 $1={\left[1,\cdots ,1\right]}_{1\times N}^{\text{T}}$， $\tilde{G}={\left[g\left(t_1\right)-g\left(t_0\right)+O\left(h\right),\cdots ,g\left(t_N\right)-g\left(t_{N-1}\right)+O\left(h\right)\right]}^{\text{T}}$，

$W^1=\left[\begin{array}{cccc}{W}_{1,0}^1& 0& \cdots & 0\\ W_{2,0}^1-W_{1,0}^1+O\left(h\right)& W_{2,1}^1& \cdots & 0\\ ⋮& ⋮& \ddots & ⋮\\ W_{N,0}^1-W_{N-1,0}^1+O\left(h\right)& W_{N,0}^1-W_{N-1,0}^1+O\left(h\right)& \cdots & W_{N,N-1}^1\end{array}\right]$，

$W^2=\left[\begin{array}{cccc}{W}_{1,0}^2& 0& \cdots & 0\\ & W_{2,1}^2& \cdots & 0\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \cdots & W_{N,N-1}^2\end{array}\right],$

而 $W_{j,j-1}^2=K_{22}\left(0\right){\displaystyle {\int }_0^1{v}_h\left(t_j+sh\right)\text{d}s+h{\displaystyle {\int }_0^1\left(1-s\right)K_{22}\left(h\left(1-s\right)\right)v_h\left(t_j+sh\right)\text{d}s}}$。

所以 ${\hat{A}}_{22}=K_{22}\left(0\right)B_N^k+O\left(h\right)$， $B_N^k$ 与(6)中的定义一致。我们可以把(10)改写为

$\left[\begin{array}{cc}I+O\left(h\right)& O\left(h\right)\\ 0& h{K}_{22}\left(0\right)B_N^k+h^{2}O\left(1\right)\end{array}\right]\left[\begin{array}{c}U\\ V\end{array}\right]=\left[\begin{array}{c}F\\ \hat{G}\end{array}\right]-\left[\begin{array}{c}{U}_0{\bar{a}}_{11}+V_0{\bar{a}}_{12}\\ U_0{\tilde{a}}_{21}+V_0{\tilde{a}}_{22}\end{array}\right],$ (12)

${\tilde{a}}_{2q}=A_{2q}\left(1:N,1:1\right),q=1,2$ 所以当 $h\to 0$ 时，方程(10)有唯一解，即方程(3)有唯一解。

接下来我们考虑配置误差 $e_h=u\left(t\right)-u_h\left(t\right)$， ${\tilde{e}}_h=v\left(t\right)-v_h\left(t\right)$。同样地，其满足方程

$\begin{array}{c}0=e_h\left(t_n\right)+h{\displaystyle \underset{j=0}{\overset{n-1}{\sum }}{\displaystyle {\int }_0^1{K}_{11}\left(t_n-\left(t_j+sh\right)\right)e_h\left(t_j+sh\right)\text{d}s}}\\ +h{\displaystyle \underset{j=0}{\overset{n-1}{\sum }}{\displaystyle {\int }_0^1{K}_{12}\left(t_n-\left(t_j+sh\right)\right){\tilde{e}}_h\left(t_j+sh\right)\text{d}s}},\end{array}$ (13)

$\begin{array}{c}0=h{\displaystyle \underset{j=0}{\overset{n}{\sum }}{\displaystyle {\int }_0^1{K}_{21}\left(t_{n+1}-\left(t_j+sh\right)\right)e_h\left(t_j+sh\right)\text{d}s}}\\ -h{\displaystyle \underset{j=0}{\overset{n-1}{\sum }}{\displaystyle {\int }_0^1{K}_{21}\left(t_n-\left(t_j+sh\right)\right)e_h\left(t_j+sh\right)\text{d}s}}\\ +h{\displaystyle \underset{j=0}{\overset{n}{\sum }}{\displaystyle {\int }_0^1{K}_{22}\left(t_{n+1}-\left(t_j+sh\right)\right){\tilde{e}}_h\left(t_j+sh\right)\text{d}s}}\\ -h{\displaystyle \underset{j=0}{\overset{n-1}{\sum }}{\displaystyle {\int }_0^1{K}_{22}\left(t_n-\left(t_j+sh\right)\right){\tilde{e}}_h\left(t_j+sh\right)\text{d}s}},\end{array}$ (14)

由Peano核定理有，当 $j=0,1,2,\cdots ,N-k-2$，在 $t_j,t_{j+1},\cdots ,t_{j+k+1}$ 处有

$e_h\left(t_j+sh\right)={\displaystyle \underset{i=0}{\overset{k+1}{\sum }}{\varphi }_i^k\left(s\right)e_h\left(t_{j+i}\right)}+O\left(h^{k+2}\right),$ (15)

${\tilde{e}}_h\left(t_j+sh\right)={\displaystyle \underset{i=0}{\overset{k+1}{\sum }}{\varphi }_i^k\left(s\right){\tilde{e}}_h\left(t_{j+i}\right)}+O\left(h^{k+2}\right),$ (16)

当 $j=N-k-1,\cdots ,N-1$ 时有

$e_h\left(t_j+sh\right)={\displaystyle \underset{i=0}{\overset{k}{\sum }}{\varphi }_i^{k-1}\left(s\right)e_h\left(t_{N-k+i}\right)}+O\left(h^{k+1}\right),$ (17)

${\tilde{e}}_h\left(t_j+sh\right)={\displaystyle \underset{i=0}{\overset{k}{\sum }}{\varphi }_i^{k-1}\left(s\right){\tilde{e}}_h\left(t_{N\text{-}k+i}\right)}+O\left(h^{k+1}\right),$ (18)

则将(15)，(16)，(17)，(18)代入(13)，(14)得到

$\left[\begin{array}{cc}I+h{A}_{11}& h{A}_{12}\\ h{\hat{A}}_{21}& h{\hat{A}}_{22}\end{array}\right]\left[\begin{array}{c}E\\ \tilde{E}\end{array}\right]=\left[\begin{array}{c}{r}_1\\ r_2\end{array}\right],$ (19)

进一步有

$\left[\begin{array}{cc}I+O\left(h\right)& O\left(h\right)\\ 0& K_{22}\left(0\right)B_N^k+O\left(h\right)\end{array}\right]\left[\begin{array}{c}E\\ \tilde{E}\end{array}\right]=\left[\begin{array}{c}{r}_1\\ {\tilde{r}}_2+r_1\end{array}\right],$ (20)

其中 $B_N^k$ 与(6)中的定义一致，则当 $N\to \infty$，由引理1、引理2可得， $B_N^k$ 有界，即

$\left[\begin{array}{cc}I+O\left(h\right)& O\left(h\right)\\ 0& K_{22}\left(0\right)B_N^k+O\left(h\right)\end{array}\right]$ 有界。

所以

${\Vert u-u_h\Vert }_{\infty }=O\left(h^{k+2}\right),$

${\Vert v-v_h\Vert }_{\infty }=O\left(h^{k+1}\right).$

证毕。

4. 数值算例

为验证理论推导，这一节我们用数值例子来支撑我们的理论结果。

例1

$\{\begin{array}{l}f\left(t\right)=u\left(t\right)+{\displaystyle {\int }_0^t\left(t-s\right)u\left(s\right)\text{d}s}+{\displaystyle {\int }_0^t{\text{e}}^{t-s}v\left(s\right)\text{d}s,}\\ g\left(t\right)={\displaystyle {\int }_0^{t}u\left(s\right)\text{d}s}+{\displaystyle {\int }_0^t{\text{e}}^{2\left(t-s\right)}v\left(s\right)\text{d}s.}\end{array}$

$t\in T=\left[0,1\right]$，该方程的解析解为 $u\left(t\right)=rt{\text{e}}^{\beta t}$， $v\left(t\right)=\cos t$。

对例1使用CBVM-k如下：

1) 对区间均匀划分： $h=\frac{T}{N}$ ；

2) 取不同的k构造插值函数： ${\varphi }_0^k\left(s\right)\cdots {\varphi }_{k+2}^k\left(s\right)$ ；

3) 离散方程，得到：

$f\left(t_n\right)=u_h\left(t_n\right)+h{\displaystyle \underset{j=0}{\overset{n-1}{\sum }}{\displaystyle {\int }_0^1\left(t_n-\left(t_j+sh\right)\right){\displaystyle \underset{i=0}{\overset{k+1}{\sum }}{\varphi }_i^k}\left(s\right)\text{d}s{U}_{j+i}}}+h{\displaystyle \underset{j=0}{\overset{n-1}{\sum }}{\displaystyle {\int }_0^1{\text{e}}^{t_n-\left(t_j+sh\right)}{\displaystyle \underset{i=0}{\overset{k+1}{\sum }}{\varphi }_i^k}\left(s\right)\text{d}s{V}_{j+i}}},$

$g\left(t_n\right)=h{\displaystyle \underset{j=0}{\overset{n-1}{\sum }}{\displaystyle {\int }_0^1{\displaystyle \underset{i=0}{\overset{k+1}{\sum }}{\varphi }_i^k}\left(s\right)\text{d}s{U}_{j+i}}}+h{\displaystyle \underset{j=0}{\overset{n-1}{\sum }}{\displaystyle {\int }_0^1{\text{e}}^{2\left(t_n-\left(t_j+sh\right)\right)}{\displaystyle \underset{i=0}{\overset{k+1}{\sum }}{\varphi }_i^k}\left(s\right)\text{d}s{V}_{j+i}}}.$

4) 整合离散方程得到： $Ax=b$ ；

5) 求解线性方程组，得到 $u\left(t\right)$， $v\left(t\right)$ 的近似解。

所有数值实验都是在MATLAB中实现的，其中积分的计算采用的是MATLAB中自带的quadgk函数。

Table 1. Absolute errors and convergence rates of for example 1 of $u\left(t\right)$ ( $r=10,\beta =-1$ )

表1. 例1中 $u\left(t\right)$ 的绝对误差和收敛阶( $r=10,\beta =-1$ )

Table 2. Absolute errors and convergence rates of for example 1 of $v\left(t\right)$ ( $r=10,\beta =-1$ )

表2. 例1中 $v\left(t\right)$ 的绝对误差和收敛阶( $r=10,\beta =-1$ )

Table 3. Absolute errors and convergence rates of for example 1 of $u\left(t\right)$ ( r = 1 , β = − 1 )

表3. 例1中 $u\left(t\right)$ 的绝对误差和收敛阶( $r=1,\beta =-1$ )

Table 4. Absolute errors and convergence rates of for example 1 of $v\left(t\right)$ ( r = 1 , β = − 1 )

表4. 例1中 $v\left(t\right)$ 的绝对误差的无穷范数和收敛阶( $r=1,\beta =-1$ )

在这里，误差表示为E与 $\tilde{E}$ 的无穷范数，收敛阶指

$\frac{lo{g}_2\left(erro{r}_{\left(i+1\right)\cdot 16}\right)-lo{g}_2\left(erro{r}_{i\cdot 16}\right)}{lo{g}_2\left(\left(i+1\right)\cdot 16\right)-lo{g}_2\left(i\cdot 16\right)},$

$i=1,2,\cdots$， $erro{r}_N$ 指相应N下的误差。

见表1~4，分别取 $k=1,2,3$，对于不同的N，结果表明 $u\left(t\right)$ 的收敛阶达到了3，4，5阶， $v\left(t\right)$ 的收敛阶达到了2，3，4阶，这一结果与我们定理2中结论一致。

5. 结论

本文介绍指标-1型积分代数方程的配置边值方法，分析了该方法的可解性和收敛性，证明了该方法求解指标-1型积分代数方程具有较高的收敛阶，数值结果证明该方法求解指标-1型积分代数方程是具有较高的收敛阶的，得到了CBVM-k求解指标-1型积分代数方程可以达到 $k+2$ 阶。

基金项目

国家自然科学资助项目(11901133)。

NOTES

^*第一作者。