Example 3. Solve the equation by the method of successive approximations

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

and estimate the error of the approximate solution.
Solution. As the zero approximation, we take $\varphi_{0}(x) \equiv 1$. Then

$$
\begin{aligned}
& \varphi_{1}(x)=\int_{0}^{1} x t^{2} \cdot 1 d t+1=1+\frac{x}{3} \\
& \varphi_{2}(x)=\int_{0}^{1} x t^{2}\left(1+\frac{t}{3}\right) d t+1=1+\frac{x}{3}\left(1+\frac{1}{4}\right) \\
& \varphi_{3}(x)=\int_{0}^{1} x t^{2}\left[1+\frac{t}{3}\left(1+\frac{1}{4}\right)\right] d t+1=1+\frac{x}{3}\left(1+\frac{1}{4}+\frac{1}{4^{2}}\right)
\end{aligned}
$$

It is not hard to see that

$$
\varphi_{m+1}(x)=1+\frac{x}{3}\left(1+\frac{1}{4}+\frac{1}{4^{2}}+\cdots+\frac{1}{4^{m}}\right) .
$$

From this

$$
\lim _{m \rightarrow \infty} \varphi_{m+1}(x)=1+\frac{4}{9} x
$$

By direct verification, we confirm that

$$
\varphi(x)=1+\frac{4}{9} x
$$

is a solution to equation (9).

To estimate the error of the $(m+1)$-th approximation, we use inequality (8). In our case,

$$
\lambda=1, \quad \Phi=1, \quad F=1, \quad A=\frac{1}{\sqrt{5}}, \quad B=\frac{1}{\sqrt{15}}
$$

so that

$$
\left|\varphi(x)-\varphi_{m+1}(x)\right| \leqslant \frac{29+\sqrt{15}}{14 \sqrt{5}(\sqrt{15})^{m}}
$$

## Problems for Independent Solution

Solve the equations using the method of successive approximations:

332. $\varphi(x)=x+4 \int_{0}^{1} x^{2} t^{2} \varphi(t) d t$.
333. $\varphi(x)=\frac{5}{6} x+\frac{1}{2} \int_{0}^{1} x t \varphi(t) d t$.
334. Find the third approximation $\varphi_{3}(x)$ to the solution of the integral equation

$$
\varphi(x)=1+\int_{0}^{1} K(x, t) \varphi(t) d t, \quad \text { where } \quad K(x, t)= \begin{cases}t, & x \geqslant t \\ x, & x \leqslant t\end{cases}
$$

and estimate the error.

Note that the main difficulty in applying the method of successive approximations lies in computing the integrals in formulas (7). Typically, approximate integration formulas must be used. Therefore, it is advisable to replace the given kernel with a degenerate one using a Taylor series expansion, and then introduce the iteration method.

## $3^{\circ}$. Integral Equations of the First Kind of Fredholm.

Suppose we have an integral equation of the first kind of Fredholm

$$
\int_{a}^{b} K(x, t) \varphi(t) d t=f(x)
$$

where the kernel $K(x, t)$ is symmetric, square-integrable, and positive definite, and $f(x) \in L_{2}(a, b)$.

Suppose it is known that equation (10) is uniquely solvable.

Then the sequence $\left\{\varphi_{n}(x)\right\}$, defined by the relation

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

where $\varphi_{0}(x) \in L_{2}(a, b)$ and $0<\lambda<2 \lambda_{1}, \lambda_{1}$ - the smallest characteristic number of the kernel $K(x, t)$, converges in the mean to the solution of equation (10).