Example. Find an approximate solution to the integral equation

$$
\varphi(x)+\int_{0}^{1} x\left(e^{x t}-1\right) \varphi(t) d t=e^{x}-x
$$
Solution. Let's take three points on the interval $[0,1]$: $x_{1}=0, x_{2}=0.5, x_{3}=1$ and substitute $x=0, x=0.5, x=1$ into equation (5). Then we get respectively

$$
\left\{\begin{array}{l}
\varphi(0)=1 \\
\varphi(0.5)+0.5 \int_{0}^{1}\left(e^{0.5 t}-1\right) \varphi(t) d t=e^{0.5}-0.5, \\
\varphi(1)+\int_{0}^{1}\left(e^{t}-1\right) \varphi(t) d t=e-1
\end{array}\right.
$$

We will replace each of the integrals with a finite sum, using Simpson's quadrature formula

$$
\int_{0}^{1} \Phi(t) d t \approx \frac{\Phi(0)+4 \Phi(0.5)+\Phi(1)}{6}
$$

For the second equation of the system (6), the integrand function is

$$
\Phi(t)=\frac{e^{0.5 t}-1}{2} \varphi(t)
$$

so that

$$
\Phi(0)=0, \quad \Phi(0.5)=\frac{e^{0.25}-1}{2} \varphi(0.5), \quad \Phi(1)=\frac{e^{0.5}-1}{2} \varphi(1)
$$

Thus,

$$
0.5 \int_{0}^{1}\left(e^{0.5 t}-1\right) \varphi(t) d t \approx \frac{e^{0.25}-1}{3} \varphi(0.5)+\frac{e^{0.5}-1}{12} \varphi(1) .
$$

For the third equation of the system (6), the integrand function is

$$
\Phi(t)=\left(e^{t}-1\right) \varphi(t)
$$

so that

$$
\Phi(0)=0, \quad \Phi(0.5)=\left(e^{0.5}-1\right) \varphi(0.5), \quad \Phi(1)=(e-1) \varphi(1),
$$

and, consequently,

$$
\int_{0}^{1}\left(e^{t}-1\right) \varphi(t) d t \approx \frac{4\left(e^{0.5}-1\right) \varphi(0.5)+(e-1) \varphi(1)}{6}
$$

Substituting (7) and (8) into (6), we get the linear system

$$
\left\{\begin{array}{l}
\varphi(0)=1 \\
\frac{e^{0.25}+2}{3} \varphi(0.5)+\frac{e^{0.5}-1}{12} \varphi(1)=e^{0.5}-0.5, \\
\frac{2\left(e^{0.5}-1\right)}{3} \varphi(0.5)+\frac{e+5}{6} \varphi(1)=e-1
\end{array}\right.
$$

or

$$
\left\{\begin{array}{l}
\varphi(0)=1 \\
1.0947 \varphi(0.5) +0.0541 \varphi(1)=1.1487 \\
0.4325 \varphi(0.5)+1.2864 \varphi(1)=1.7183
\end{array}\right.
$$

Solving system (9), we get

$$
\varphi(0)=1, \quad \varphi(0.5)=0.9999, \quad \varphi(1)=0.9996 .
$$

We take the approximate solution of equation (5) to be the function

$$
\tilde{\varphi}(x)=e^{x}-x-\sum_{m=1}^{3} A_{m} x\left(e^{x x_{m}}-1\right) \varphi\left(x_{m}\right)
$$

where $x_{1}=0, x_{2}=0.5, x_{3}=1, A_{1}=\frac{1}{6}, A_{2}=\frac{2}{3} A_{3}=\frac{1}{6}$, or

$$
\tilde{\varphi}(x)=e^{x}-x\left(0.6666 e^{0.5 x}+0.1666 e^{x}\right)-0.1668 x .
$$

The exact solution of this integral equation is $\varphi(x) \equiv 1$.

## Problems for Independent Solution

Solve the following integral equations using the replacement of the integral with a finite sum:

316. $\varphi(x)+\int_{0}^{1} x e^{x t} \varphi(t) d t=e^{x}$.
317. $\varphi(x)-\int_{0}^{1} \frac{x+t}{1+x+t} \varphi(t) d t=\ln \frac{2+x}{1+x}$.
318. $\varphi(x)+\pi \int_{0}^{1} x^{2} \cos \pi x t \cdot \varphi(t) d t=\pi x(1+\sin \pi x)-2 \sin ^{2} \frac{\pi x}{2}$.

## § 25. Method of Successive Approximations

$1^{\circ}$. Volterra integral equations of the second kind. Suppose we have a Volterra integral equation of the second kind

$$
\varphi(x)=f(x)+\lambda \int_{0}^{x} K(x, t) \varphi(t) d t
$$

We will assume that $f(x)$ is continuous on $[0, a]$, and the kernel $K(x, t)$ is continuous for $0 \leqslant x \leqslant a, 0 \leqslant t \leqslant x$.

Take any continuous function $\varphi_{0}(x)$ on $[0, a]$. Substituting $\varphi_{0}(x)$ for $\varphi(x)$ in the right-hand side of equation (1), we get

$$
\varphi_{1}(x)=f(x)+\lambda \int_{0}^{x} K(x, t) \varphi_{0}(t) d t
$$

The function $\varphi_{1}(x)$ thus defined is also continuous on the interval $[0, a]$. Continuing this process, we obtain a sequence of functions

$$
\varphi_{0}(x), \varphi_{1}(x), \ldots, \varphi_{n}(x), \ldots
$$

where

$$
\varphi_{n}(x)=f(x)+\lambda \int_{0}^{x} K(x, t) \varphi_{n-1}(t) d t
$$

Under the assumptions made about $f(x)$ and $K(x, t)$, the sequence $\left\{\varphi_{n}(x)\right\}$ converges as $n \rightarrow \infty$ to the solution $\varphi(x)$ of the integral equation (1).

If, in particular, we take $\varphi_{0}(x)$ to be $f(x)$, then $\varphi_{n}(x)$ will be the partial sums of the series (2) from §, which defines the solution of the integral equation (1). A good choice of the "zero" approximation $\varphi_{0}(x)$ can lead to rapid convergence of the sequence $\left\{\varphi_{n}(x)\right\}$ to the solution of the integral equation.