11.3.6 迭代法

为了解积分方程

$$\begin{array}{}\text{(11.54a)}& f\left(x\right)={\int }_a^{b}K\left(x,y\right)\phi \left(y\right)\mathrm{d}y\left(c\le x\le d\right),\end{array}$$

从 ${\alpha }_0\left(x\right)=f\left(x\right)$ 开始,对于 $n=1,2,\cdots$ 确定诸函数

$$\begin{array}{}\text{(11.54b)}& {\beta }_n\left(y\right)={\int }_c^{d}K\left(x,y\right){\alpha }_{n-1}\left(x\right)\mathrm{d}x,\end{array}$$$$\begin{array}{}\text{(11.54c)}& {\alpha }_n\left(x\right)={\int }_a^{b}K\left(x,y\right){\beta }_n\left(y\right)\mathrm{d}y.\end{array}$$

如果 (11.54a) 存在一个平方可积解 $\phi \left(y\right)$ ,则下列等式成立:

$${\int }_a^b\phi \left(y\right){\beta }_n\left(y\right)\mathrm{d}y={\int }_a^b{\int }_c^d\phi \left(y\right)K\left(x,y\right){\alpha }_{n-1}\left(x\right)\mathrm{d}x\mathrm{d}y$$$$\begin{array}{}\text{(11.54d)}& ={\int }_c^{d}f\left(x\right){\alpha }_{n-1}\left(x\right)\mathrm{d}x\left(n=1,2,\cdots \right).\end{array}$$

函数系 (11.54b),(11.54c) 的正交化和规范化给出了规范正交系 $\left({\alpha }_n^{\ast }\left(x\right)\right)$ 和 $\left({\beta }_n^{\ast }\left(y\right)\right)$ . 利用施密特正交化方法,则 ${\beta }_n^{\ast }\left(y\right)$ 有形式

$$\begin{array}{}\text{(11.54e)}& {\beta }_n^{\ast }\left(y\right)=\sum _{j=1}^n{b}_{nj}{\beta }_j\left(y\right)\left(n=1,2,\cdots \right).\end{array}$$

现在假设(11.54a)的解 $\phi \left(y\right)$ 有级数表达式 ${}^{\text{①}}$

$$\begin{array}{}\text{(11.54f)}& \phi \left(y\right)=\sum _{n=1}^{\mathrm{\infty }}{c}_n{\beta }_n^{\ast }\left(y\right).\end{array}$$

注意到(11.54d),对于系数 $c_n$ 即有

$$c_n={\int }_a^b\phi \left(y\right){\beta }_n^{\ast }\left(y\right)\mathrm{d}y=\sum _{j=1}^n{b}_{nj}{\int }_a^b\phi \left(y\right){\beta }_j\left(y\right)\mathrm{d}y=\sum _{j=1}^n{b}_{nj}{\int }_c^{d}f\left(x\right){\alpha }_{j-1}\left(x\right)\mathrm{d}x.$$

(11.54g)

为了有形如(11.54f)的解,下述两个条件都是充要条件:

(1) ${\int }_c^d{\left[f\left(x\right)\right]}^2\mathrm{d}x=\sum _{n=1}^{\mathrm{\infty }}{\left|{\int }_c^{d}f\left(x\right){\alpha }_n^{\ast }\left(x\right)\mathrm{d}x\right|}^2$ ;(11.55a)

(2) $\sum _{n=1}^{\mathrm{\infty }}{\left|c_n\right|}^2<\mathrm{\infty }$ .(11.55b)