Example 1. Solve the integral equation by the method of successive approximations

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

taking $\varphi_{0}(x) \equiv 0$.
Solution. Since $\varphi_{0}(x) \equiv 0$, then $\varphi_{1}(x)=1$. Further,

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

Obviously,

\[
\varphi_{n}(x)=1+\frac{x}{1!}+\frac{x^{2}}{2!}+\ldots+\frac{x^{n-1}}{(n-1)!}
\]

Thus, $\varphi_{n}(x)$ is the $n$-th partial sum of the series $\sum_{n=0}^{\infty} \frac{x^{n}}{n!}=e^{x}$. Therefore, $\varphi_{n}(x) \underset{n \rightarrow \infty}{\longrightarrow} e^{x}$. It is not difficult to verify that the function $\varphi(x)=e^{x}$ is a solution to the given integral equation.

$\square$

## Problems for Independent Solution

Solve the following integral equations using the method of successive approximations:

319. $\varphi(x)=x-\int_{0}^{x}(x-t) \varphi(t) d t, \quad \varphi_{0}(x) \equiv 0$.
320. $\varphi(x)=1-\int_{0}^{x}(x-t) \varphi(t) d t, \quad \varphi_{0}(x) \equiv 0$.
321. $\varphi(x)=1+\int_{0}^{x}(x-t) \varphi(t) d t, \quad \varphi_{0}(x)=1$.
322. $\varphi(x)=x+1-\int_{0}^{x} \varphi(t) d t$, a) $\varphi_{0}(x)=1$, b) $\varphi_{0}(x)=x+1$.
323. $\varphi(x)=\frac{x^{2}}{2}+x-\int_{0}^{x} \varphi(t) d t$, a) $\varphi_{0}(x)=1$, b) $\varphi_{0}(x)=x$, c) $\varphi_{0}(x)=\frac{x^{2}}{2}+x$.
324. $\varphi(x)=1+x+\int_{0}^{x}(x-t) \varphi(t) d t, \quad \varphi_{0}(x)=1$.
325. $\varphi(x)=2 x+2-\int_{0}^{x} \varphi(t) d t, \quad$ a) $\varphi_{0}(x)=1, \quad$ b) $\varphi_{0}(x)=2$.
326. $\varphi(x)=2 x^{2}+2-\int_{0}^{x} x \varphi(t) d t$, a) $\varphi_{0}(x)=2, \quad$ b) $\varphi_{0}(x)=2 x$.
327. $\varphi(x)=\frac{x^{3}}{3}-2 x-\int_{0}^{x} \varphi(t) d t, \quad \varphi_{0}(x)=x^{2}$.
328. Let $K(x, t)$ satisfy the condition

\[
\int_{0}^{a} \int_{0}^{x} K^{2}(x, t) d t d x<+\infty
\]

Prove that the equation

\[
\varphi(x)-\lambda \int_{0}^{x} K(x, t) \varphi(t) d t=0
\]

has a unique solution $\varphi(x) \equiv 0$ in the class $L_{2}(0, a)$ for any $\lambda$.

The method of successive approximations can also be applied to solving nonlinear Volterra integral equations of the form

\[
y(x)=y_{0}+\int_{0}^{x} F[t, y(t)] d t
\]

or more generally,

\[
\varphi(x)=f(x)+\int_{0}^{x} F[x, t, \varphi(t)] d t
\]

under quite broad assumptions about the functions $F(x, t, z)$ and $f(x)$. The equation of the form (2) arises from the problem of solving the differential equation

\[
\frac{d y}{d x}=F(x, y),\left.\quad y\right|_{x=0}=y_{0}
\]

As in the case of linear integral equations, we seek the solution of equation (3) as the limit of the sequence $\left\{\varphi_{n}(x)\right\}$, where, for example, $\varphi_{0}(x)=f(x)$, and the subsequent elements $\varphi_{k}(x)$ are computed sequentially by the formula

\[
\varphi_{k}(x)=f(x)+\int_{0}^{x} F\left(x, t, \varphi_{k-1}(t)\right) d t \quad(k=1,2, \ldots)
\]

If $f(x)$ and $F(x, t, z)$ are square-integrable and satisfy the conditions

\[
\begin{aligned}
\left|F\left(x, t, z_{2}\right)-F\left(x, t, z_{1}\right)\right| & \leqslant a(x, t)\left|z_{2}-z_{1}\right| \\
\left|\int_{0}^{x} F(x, t, f(t)) d t\right| & \leqslant n(x)
\end{aligned}
\]

where the functions $a(x, t)$ and $n(x)$ are such that in the main domain ($0 \leqslant t \leqslant x \leqslant a$)

\[
\int_{0}^{a} n^{2}(x) d x \leqslant N^{2}, \quad \int_{0}^{a} d x \int_{0}^{x} a^{2}(x, t) d t \leqslant A^{2}
\]

then the nonlinear Volterra integral equation of the second kind (3) has a unique solution $\varphi(x) \in L_{2}(0, a)$, which is defined as the limit $\varphi_{n}(x)$ as $n \rightarrow \infty$:

\[
\varphi(x)=\lim _{n \rightarrow \infty} \varphi_{n}(x),
\]

where the functions $\varphi_{n}(x)$ are found using the recursive formulas (4). As $\varphi_{0}(x)$, one can take any function from $L_{2}(0, a)$ (in particular, a continuous function) for which the condition (5) is satisfied. Note that a successful choice of the initial approximation can simplify the solution of the integral equation.