基于泛函修正平均法的第二类积分方程的改进迭代法An Improved Iterative Method for the Second Kind of Integral Equation Based on Modified Functional Averaging Method

Abstract: In this paper, the modified functional average method is extended to the second kind of linear Volterra Integral Equation, and the error of the method is analyzed. Then the first iteration of the method is adjusted, that is, in the iterative solution of u1, the modified term which is in the in-complete substitution form of u0n-1(t)(u0(t)+a1) replaces the nonlinear term un(t). Then the method is applied to the solution of a special form of nonlinear Fredholm Integral Equation. Finally, the feasibility and effectiveness of the method are verified by numerical examples in this paper.

1. 引言

2. 准备知识

2.1. 压缩映射与不动点定理

$u\left(x\right)=\lambda {\int }_{a}^{x}k\left(x,t\right)\psi \left(t,u\left(t\right)\right)\text{d}t+f\left(x\right),\text{}x\in \left[a,b\right],\text{}t\in \left[a,b\right],$ (1)

$u\left(x\right)=\lambda {\int }_{a}^{x}k\left(x,t\right)u\left(t\right)\text{d}t+f\left(x\right),\text{}x\in \left[a,b\right],\text{}t\in \left[a,b\right],$ (2)

$u\left(x\right)=\lambda {\int }_{a}^{b}k\left(x,t\right){u}^{m}\left(t\right)\text{d}t+f\left(x\right),\text{}x\in \left[a,b\right],\text{}t\in \left[a,b\right],\text{}m\in {Z}^{+},\text{}m\ne 1.$ (3)

2.2. 解的存在唯一性

$F:C\left[a,b\right]\to C\left[a,b\right],\forall u\in C\left[a,b\right],\exists \xi \left(t\right)\in C\left[a,b\right]$，有

$F\left(u\left(x\right)\right)=\lambda {\int }_{0}^{x}k\left(x,t\right)\psi \left(t,u\left(t\right)\right)\text{d}t+f\left(x\right).$

1) $\psi \left(t,u\right)\in {C}^{1}\left(\left[a,b\right]×C\left[a,b\right]\right)$

2) $0\le q=|\lambda |‖{\int }_{0}^{x}|k\left(x,t\right)||{\psi }_{u}\left(t,\xi \left(t\right)\right)|\text{d}t‖<1$

$F\left({u}_{1}\left(x\right)\right)-F\left({u}_{2}\left(x\right)\right)=\lambda {\int }_{0}^{x}k\left(x,t\right)\left(\psi \left(t,{u}_{1}\left(t\right)\right)-\psi \left(t,{u}_{2}\left(t\right)\right)\right)\text{d}t.$

$\psi \left(t,{u}_{1}\left(t\right)\right)-\psi \left(t,{u}_{2}\left(t\right)\right)={\psi }_{u}\left(t,\xi \left(t\right)\right)\left({u}_{1}\left(t\right)-{u}_{2}\left(t\right)\right),$

$\begin{array}{c}‖F{u}_{1}-F{u}_{2}‖\le |\lambda |{\int }_{0}^{x}|k\left(x,t\right)|‖{\psi }_{u}\left(t,\xi \left(t\right)\right)\left({u}_{1}\left(t\right)-{u}_{2}\left(t\right)\right)\text{d}t‖\\ \le |\lambda |‖\text{}{\int }_{0}^{x}|k\left(x,t\right)\psi \left(t,\xi \left(t\right)\right)|\text{d}t‖\cdot ‖{u}_{1}-{u}_{2}‖\triangleq q‖{u}_{1}-{u}_{2}‖.\end{array}$

2.3. 皮卡尔迭代法求解积分方程

${u}_{0}\left(x\right)=f\left( x \right)$

${u}_{n+1}\left(x\right)=f\left(x\right)+\lambda {\int }_{0}^{x}k\left(x,t\right)\psi \left(t,{u}_{n}\left(t\right)\right)\text{d}t,\text{}n=0,1,2,\cdots .$

$u\left(x\right)=\underset{n\to \infty }{\mathrm{lim}}{u}_{n}\left(x\right).$

(a) 方程 $u=Fu$ 有唯一精确解等价于F有唯一不动点 ${u}^{\ast }\in X$

(b) 对于任意初始值 ${u}_{0}\in X$，迭代逼近序列 ${u}_{n+1}=F{u}_{n},n\in N$ 是收敛于 ${u}^{\ast }$ 的；

(c) $\forall n\in N$，使得皮卡尔迭代法的解有如下误差估计

$‖{u}_{n+1}-{u}_{n}‖\le {q}^{n}‖{u}_{1}-{u}_{0}‖,$ (4)

$‖{u}_{n}-{u}^{\ast }‖\le \frac{{q}^{n}}{1-q}‖{u}_{1}-{u}_{0}‖.$ (5)

(a) 结合定理2及定理3显然(a)成立。

(b) 由条件2

$0\le q=|\lambda |‖{\int }_{0}^{x}|K\left(x,t\right)||{\psi }_{u}\left(t,\xi \left(t\right)\right)|\text{d}t‖<1,$

$‖{u}_{n}-{u}^{\ast }‖=‖F{u}_{n-1}-F{u}^{\ast }‖\le q‖{u}_{n-1}-{u}^{\ast }‖\le \cdot \cdot \cdot \le {q}^{n}‖{u}_{0}-{u}^{\ast }‖,$

$‖{u}_{n}-{u}^{\ast }‖\to 0,$

(c) 由(b)得

$‖{u}_{n+1}-{u}_{n}‖=‖F{u}_{n}-F{u}_{n-1}‖\le q‖{u}_{n}-{u}_{n-1}‖,\text{}n=1,2,3,\cdots .$

$‖{u}_{n+1}-{u}_{n}‖\le {q}^{n}‖{u}_{1}-{u}_{0}‖.$

$\forall p\in N$ 可得

$\begin{array}{c}‖{u}_{n+p}-{u}_{n}‖\le ‖{u}_{n+p}-{u}_{n+p-1}‖+‖{u}_{n+p-1}-{u}_{n+p-2}‖+\cdot \cdot \cdot +‖{u}_{n+1}-{u}_{n}‖\\ \le \left({q}^{n+p-1}+{q}^{n+p-2}+\cdot \cdot \cdot +{q}^{n}\right)\cdot ‖{u}_{1}-{u}_{0}‖\le \frac{{q}^{n}}{1-q}‖{u}_{1}-{u}_{0}‖.\end{array}$

$p\to \infty$，则 ${u}_{n+p}\to {u}^{\ast }$，由(b)则可得(5)式成立。

3. 改进的皮卡尔迭代法

3.1. 第二类线性积分方程的改进皮卡尔迭代法

$u\left(x\right)=\lambda {\int }_{a}^{x}k\left(x,t\right)u\left(t\right)\text{d}t+f\left(x\right),\text{}x\in \left[a,b\right],\text{}t\in \left[a,b\right],$

$\xi \in \left(a,x\right]$，则

${\alpha }_{n}=\frac{1}{\xi -a}{\int }_{a}^{\xi }{\delta }_{n}\left(x\right)\text{d}x,$

$u\left(\xi \right)=\lambda {\int }_{a}^{\xi }k\left(\xi ,t\right)u\left(t\right)\text{d}t+f\left(\xi \right),\text{}\xi \in \left(a,x\right],\text{}t\in \left[a,b\right],$

${u}_{n}\left(x\right)=\lambda {\int }_{a}^{x}k\left(x,t\right)\left[{u}_{n-1}\left(t\right)+{\alpha }_{n}\right]\text{d}t+f\left(x\right),\text{}x\in \left[a,b\right],\text{}t\in \left[a,b\right],$

${\delta }_{n}\left(\xi \right)={u}_{n}\left(\xi \right)-{u}_{n-1}\left(\xi \right),\text{}n=1,2,\cdots .$ (6)

$C\left(\lambda \right)=\left(\xi -a\right)-\lambda {\int }_{a}^{\xi }{\int }_{a}^{\xi }k\left(x,t\right)\text{d}x\text{d}t,$ (7)

$\begin{array}{c}{\alpha }_{n}=\frac{1}{\xi -a}{\int }_{a}^{\xi }{\delta }_{n}\left(x\right)\text{d}x=\frac{1}{\xi -a}{\int }_{a}^{\xi }\left[{u}_{n}\left(x\right)-{u}_{n-1}\left(x\right)\right]\text{d}x\\ =\frac{1}{\xi -a}{\int }_{a}^{\xi }\left(\lambda {\int }_{a}^{\xi }k\left(x,t\right)\left[{\delta }_{n-1}\left(t\right)-{\alpha }_{n-1}+{\alpha }_{n}\right]\right)\text{d}x\\ =\frac{\lambda }{C\left(\lambda \right)}{\int }_{a}^{\xi }{\int }_{a}^{\xi }k\left(x,t\right)\left[{\delta }_{n-1}\left(t\right)-{\alpha }_{n-1}\right]\text{d}t\text{d}x.\end{array}$ (8)

${\alpha }_{0}=0,\text{}{\delta }_{0}\left(x\right)={u}_{0}\left(x\right),$

${u}_{n}\left(\xi \right)=f\left(\xi \right)+\lambda {\int }_{a}^{\xi }k\left(\xi ,t\right)\left[{u}_{n-1}\left(t\right)+{\alpha }_{n}\right]\text{d}t.$

$\xi \in \left(a,x\right]$ 任意性，上式可得

${u}_{n}\left(x\right)=f\left(x\right)+\lambda {\int }_{a}^{x}k\left(x,t\right)\left[{u}_{n-1}\left(t\right)+{\alpha }_{n}\right]\text{d}t.$ (9)

$\xi =a$，则

$u\left(x\right)=f\left(0\right).$

3.2. 形如(3)式型非线性积分方程的改进的皮卡尔迭代法

$u\left(x\right)=\lambda {\int }_{a}^{b}k\left(x,t\right){u}^{m}\left(t\right)\text{d}t+f\left(x\right),\text{}x\in \left[a,b\right],\text{}t\in \left[a,b\right],\text{}m\in {Z}^{+},\text{}m\ne 1,$

${u}_{0}\left(x\right)=f\left(x\right),$

$u\left(x\right)=\lambda {\int }_{a}^{b}k\left(x,t\right){u}_{0}^{m-1}\left(t\right)u\left(t\right)\text{d}t+f\left(x\right),\text{}t\in \left[a,b\right],\text{}m\in {Z}^{+},\text{}m\ne 1$ (10)

${\delta }_{1}\left(x\right)={u}_{1}\left(x\right)-{u}_{0}\left(x\right),\text{}n=1,2,\cdots .$

${C}_{1}\left(\lambda \right)=\left(b-a\right)-\lambda {\int }_{a}^{b}{\int }_{a}^{b}k\left(x,t\right){u}_{0}^{m-1}\left(t\right)\text{d}x\text{d}t,$

$\begin{array}{c}{\alpha }_{1}=\frac{1}{b-a}{\int }_{a}^{b}{\delta }_{1}\left(x\right)\text{d}x=\frac{1}{b-a}{\int }_{a}^{b}\left[{u}_{1}\left(x\right)-{u}_{0}\left(x\right)\right]\text{d}x\\ =\frac{1}{b-a}{\int }_{a}^{b}\left(\lambda {\int }_{a}^{b}K\left(x,t\right)\left[{\delta }_{1}\left(t\right)-{\alpha }_{0}+{\alpha }_{1}\right]\right)\text{d}x\\ =\frac{\lambda }{C\left(\lambda \right)}{\int }_{a}^{b}{\int }_{a}^{b}k\left(x,t\right){u}_{0}^{m-1}\left(t\right)\left[{\delta }_{0}\left(t\right)-{\alpha }_{0}\right]\text{d}t\text{d}x.\end{array}$

${u}_{1}\left(x\right)=f\left(x\right)+\lambda {\int }_{a}^{b}k\left(x,t\right){u}_{0}^{m-1}\left(t\right)\left[{u}_{0}\left(t\right)+{\alpha }_{1}\right]\text{d}t.$

${u}_{n+1}\left(x\right)=f\left(x\right)+\lambda {\int }_{0}^{b}k\left(x,t\right){u}_{n}^{m}\left(x\right)\text{d}t,\text{}\left(n\ge 1\right).$

4. 数值模拟

$u\left(x\right)={\int }_{0}^{x}\left(x-t\right)u\left(t\right)\text{d}t+f\left(x\right),$

$\xi =0$，由3.1节方法

$u\left(0\right)=f\left(0\right),$

$\xi \in \left(0,x\right]$，有如下等式成立

$u\left(\xi \right)={\int }_{0}^{\xi }\left(\xi -t\right)u\left(t\right)\text{d}t+f\left(\xi \right),\text{}\xi \in \left(0,x\right]\text{,}$

${u}_{n}\left(\xi \right)=f\left(\xi \right)+{\int }_{a}^{\xi }k\left(\xi ,t\right)\left[{u}_{n-1}\left(t\right)+{\alpha }_{n}\right]\text{d}t.$

$f\left(x\right)$ 为0次近似解

${u}_{0}\left(x\right)=1+x,\text{\hspace{0.17em}}\text{\hspace{0.17em}}C\left(\lambda \right)=\xi -{\int }_{0}^{\xi }{\int }_{0}^{\xi }\left(x-t\right)\text{d}x\text{d}t=\xi ,$

${\alpha }_{1}=\frac{1}{\xi }{\int }_{0}^{\xi }{\int }_{0}^{\xi }\left(x-t\right)\left(1+t\right)\text{d}x\text{d}t=-\frac{1}{12}{\xi }^{3},$

${u}_{1}\left(\xi \right)=1+\xi +{\int }_{0}^{\xi }\left(\xi -t\right)\left(1+t-\frac{1}{12}{\xi }^{3}\right)\text{d}t=1+\xi +\frac{{\xi }^{2}}{2}+\frac{{\xi }^{3}}{3!}-\frac{{\xi }^{5}}{4!}.$

$\xi$ 的任意性可知 ${u}_{1}\left(x\right)=1+x+\frac{{x}^{2}}{2}+\frac{{x}^{3}}{3!}-\frac{{x}^{5}}{4!}$，可以发现前4项与解析解泰勒展开一致，可见逼近效果较好。

$u\left(x\right)=\lambda {\int }_{0}^{x}\sqrt{x}\left(t+10\right)u\left(t\right)\text{d}t+f\left(x\right),$ (11)

$x=1$，取 $\lambda =\frac{1}{10},f\left(x\right)=\frac{22}{75}\sqrt{x}$，可得对应方程解析解 $u\left(x\right)=\sqrt{x}$

$f\left(x\right)$ 为零次迭代解，有

${u}_{0}\left(x\right)=\frac{22}{75}\sqrt{x},$

$C\left(\lambda \right)=1-\frac{1}{10}{\int }_{0}^{1}{\int }_{0}^{1}\sqrt{x}\left(t+10\right)\text{d}x\text{d}t=\frac{3}{10},$

${\alpha }_{1}=\frac{10}{3}×\frac{1}{10}{\int }_{0}^{1}{\int }_{0}^{1}\frac{22}{75}\sqrt{x}\left(t+10\right)\sqrt{t}\text{d}x\text{d}t=\frac{4664}{10125},$

${u}_{1}\left(x\right)=\frac{22}{75}\sqrt{x}+\frac{1}{10}{\int }_{0}^{1}\left(\frac{22}{75}\sqrt{t}+\frac{4664}{10125}\right)\sqrt{x}\left(t+10\right)\text{d}t\approx 0.9843\sqrt{x}.$

${\delta }_{1}\left(x\right)={u}_{1}\left(x\right)-{u}_{0}\left(x\right)\approx \frac{6910}{100000}\sqrt{x},$

${\alpha }_{2}=\frac{10}{3}×\frac{1}{10}{\int }_{0}^{1}{\int }_{0}^{1}\left(\frac{6910}{10000}\sqrt{t}-\frac{4664}{10125}\right)\sqrt{x}\left(t+10\right)\text{d}x\text{d}t\approx 0.0103,$

${u}_{2}\left(x\right)=\frac{22}{75}\sqrt{x}+\frac{1}{10}{\int }_{0}^{1}\left(\frac{9843}{10000}\sqrt{t}-\frac{103}{10000}\right)\sqrt{x}\left(t+10\right)\text{d}t\approx 0.9997\sqrt{x}.$

${\delta }_{2}\left(x\right)=\frac{154}{1000}\sqrt{x}\text{,}{\alpha }_{\text{3}}=\frac{15037}{100000},$

${u}_{3}\left(x\right)=\frac{22}{75}\sqrt{x}+\frac{1}{10}{\int }_{0}^{1}\left(\frac{9997\sqrt{t}}{10000}+\frac{15037}{100000}\right)\sqrt{x}\left(t+10\right)\text{d}t\approx 0.99995\sqrt{x}.$

Table 1. Comparison of approximate and analytical solutions of equation (11) under Picard iterative method and improved Picard iterative method

$RE=\frac{\sqrt{\underset{j=0}{\overset{10}{\sum }}{|u\left({x}_{j}\right)-{u}_{n}\left({x}_{j}\right)|}^{2}}}{\sqrt{\underset{j=0}{\overset{10}{\sum }}{|u\left({x}_{j}\right)|}^{2}}}.$ (12)

$u\left(x\right)=\lambda {\int }_{0}^{1}xt{u}^{m}\left(t\right)\text{d}t+f\left(x\right).$

$u\left(x\right)={\int }_{0}^{1}xt{u}^{2}\left(t\right)\text{d}t+f\left(x\right),$ (13)

${u}_{0}\left(x\right)=1,$

${u}_{1}\left(x\right)=1+0.083x,\text{}{u}_{2}\left(x\right)=1+0.140x,\text{}{u}_{3}\left(x\right)=1+0.182x,\text{\hspace{0.17em}}\cdots ,\text{}{u}_{7}\left(x\right)=1+0.269x,\text{\hspace{0.17em}}\cdots .$

$u\left(x\right)={\int }_{0}^{1}xtu\left(t\right)\text{d}t+1-\frac{5}{12}x.$

$\lambda =1$，可得

$C\left(\lambda \right)=\frac{3}{4},$

${\alpha }_{1}=\frac{4}{3}{\int }_{0}^{1}{\int }_{0}^{1}xt\text{d}x\text{d}t=\frac{1}{3},$

${u}_{1}\left(x\right)=1-\frac{5}{12}x+{\int }_{0}^{1}xt\left(1+\frac{1}{3}\right)\text{d}t=1+\frac{x}{4}.$

Table 2. Comparison of approximate solutions and analytical solutions obtained by taking different initial iterative solutions between Picard iterative method and improved Picard iterative method for equation (13)

$u\left(x\right)=\frac{1}{2}{\int }_{0}^{1}xt{u}^{2}\left(t\right)\text{d}t+f\left(x\right),$ (14)

Table 3. Comparison of approximate solution and exact solution of equation (14) under Picard iterative method and improved Picard iterative method

5. 结束语

NOTES

*通讯作者。

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