Example 2. Solve the integral equation by the method of successive approximations

$$
\varphi(x)=\int_{0}^{x} \frac{1+\varphi^{2}(t)}{1+t^{2}} d t
$$

taking as the zeroth approximation: 1) $\varphi_{0}(x)=0$; 2) $\varphi_{0}(x)=x$.
Solution. 1) Let $\varphi_{0}(x)=0$. Then

\[
\begin{aligned}
& \varphi_{1}(x)=\int_{0}^{x} \frac{d t}{1+t^{2}}=\operatorname{arctg} x \\
& \varphi_{2}(x)=\int_{0}^{x} \frac{1+\operatorname{arctg}^{2} t}{1+t^{2}} d t=\operatorname{arctg} x+\frac{1}{3} \operatorname{arctg}^{3} x \\
& \varphi_{3}(x)=\int_{0}^{x} \frac{1+\left(\operatorname{arctg} t+\frac{1}{3} \operatorname{arctg}^{3} t\right)^{2}}{1+t^{2}} d t= \\
& =\operatorname{arctg} x+\frac{1}{3} \operatorname{arctg}^{3} x+\frac{2}{3 \cdot 5} \operatorname{arctg}^{5} x+\frac{1}{7 \cdot 9} \operatorname{arctg}^{7} x, \\
& \varphi_{4}(x)=\int_{0}^{x} \frac{1+\varphi_{3}^{3}(t)}{1+t^{2}} d t=\operatorname{arctg} x+\frac{1}{3} \operatorname{arctg}^{3} x+\frac{2}{3 \cdot 5} \operatorname{arctg}^{3} x+ \\
& +\frac{17}{5 \cdot 7 \cdot 9} \operatorname{arctg}^{7} x+\frac{38}{5 \cdot 7 \cdot 9^{2}} \operatorname{arctg}^{9} x+\frac{134}{9 \cdot 11 \cdot 21 \cdot 25} \operatorname{arctg}^{11} x+ \\
& +\frac{4}{3 \cdot 5 \cdot 7 \cdot 9 \cdot 13} \operatorname{arctg}^{13} x+\frac{1}{7^{2} \cdot 9^{2} \cdot 15} \operatorname{arctg}^{15} x
\end{aligned}
\]

Denoting $\operatorname{arctg} x=u$ and comparing the expressions for $\varphi_{n}(x)$ with the expansion

\[
\operatorname{tg} u=\sum_{\nu=1}^{\infty}(-1)^{\nu-1} \frac{2^{2 \nu}\left(2^{2 \nu}-1\right)}{(2 \nu)!} B_{2 \nu} u^{2 \nu-1}, \quad|u|<\frac{\pi}{2}
\]

where $B_{v}$ - Bernoulli numbers ${ }^{1)}$, we notice that

\[
\varphi_{n}(x) \underset{n \rightarrow \infty}{\longrightarrow} \operatorname{tg}(\operatorname{arctg} x)=x
\]

It is not difficult to verify that the function $\varphi(x)=x$ is a solution to the given integral equation.

2) Let $\varphi_{0}(x)=0$. Then

\[
\varphi_{1}(x)=\int_{0}^{x} \frac{1+t^{2}}{1+t^{2}} d t=x
\]

Similarly, we find $\varphi_{n}(x)=x(n=2,3, \ldots)$.

${ }^{1)}$ Bernoulli numbers $B_{2 \nu+1}$ with odd indices are all zero, except for $B_{1}=-\frac{1}{2}$. The number
$B_{0}=1$, and the numbers $B_{2 \nu}$ are determined by the recurrence formulas

\[
B_{2 \nu}=-\frac{1}{2 \nu+1}+\frac{1}{2}-\sum_{k=2}^{2 \nu-2} \frac{2 \nu(2 \nu-1) \ldots(2 \nu-2 k+2)}{k!} B_{k}
\]

Thus, the sequence $\left\{\varphi_{n}(x)\right\}$ is a stationary sequence $\{x\}$, the limit of which is $\varphi(x)=x$. The solution to the given integral equation is obtained immediately:

\[
\varphi(x)=x
\]

$\Delta$

## Problems for Independent Solution

329. Solve the integral equation by the method of successive approximations

\[
\varphi(x)=\int_{0}^{x} \frac{t \varphi(t)}{1+t+\varphi(t)} d t
\]

330. Find the second approximation $\varphi_{2}(x)$ of the solution to the integral equation by the method of successive approximations

\[
\varphi(x)=1+\int_{0}^{x}\left[\varphi^{2}(t)+t \varphi(t)+t^{2}\right] d t
\]

331. Find the third approximation $\varphi_{3}(x)$ of the solution to the integral equation by the method of successive approximations

\[
\varphi(x)=\int_{0}^{x}\left[t \varphi^{2}(t)-1\right] d t
\]

$2^{\circ}$. Fredholm Integral Equations of the Second Kind. Suppose we have a Fredholm integral equation of the second kind

\[
\varphi(x)=f(x)+\lambda \int_{a}^{b} K(x, t) \varphi(t) d t
\]

We construct a sequence of functions $\left\{\varphi_{n}(x)\right\}$ using the recurrence formula

\[
\varphi_{n}(x)=f(x)+\lambda \int_{a}^{b} K(x, t) \varphi_{n-1}(t) d t
\]

The functions $\varphi_{n}(x)(n=1,2, \ldots)$ are considered as approximations to the desired solution of the equation, and the zeroth approximation $\varphi_{0}(x)$ can be chosen arbitrarily.

If the condition

\[
|\lambda|<\frac{1}{B}, \quad \text { where } \quad B=\sqrt{\int_{a}^{b} \int_{a}^{b} K^{2}(x, t) d x d t}
\]

is satisfied, then the sequence (7) converges to the solution of equation (6) in the metric $L_{2}(a, b)$. The magnitude of the error of the $(m+1)$-th approximation is determined by the inequality

\[
\left|\varphi(x)-\varphi_{m+1}(x)\right| \leqslant A B^{-1}|\lambda B|^{m+1}\left(\Phi+\frac{F}{1-|\lambda B|}\right)
\]

where

\[
F=\sqrt{\int_{a}^{b} f^{2}(x) d x}, \quad \Phi=\sqrt{\int_{a}^{b} \varphi_{0}^{2}(x) d x}, \quad A=\sqrt{\max _{a \leqslant x \leqslant b} \int_{a}^{b} K^{2}(x, t) d t} .
\]