Example 1. Find the approximate value of the smallest characteristic number of the kernel by the Ritz method

$$
K(x, t)=x t ; \quad a=0, b=1
$$
Solution. As the coordinate system of functions $\psi_{n}(x)$, we choose the system of Legendre polynomials: $\psi_{n}(x)=P_{n}(2 x-1)$. In formula (1), we limit ourselves to two terms, so that

$$
\varphi_{2}(x)=a_{1} \cdot P_{0}(2 x-1)+a_{2} \cdot P_{1}(2 x-1) .
$$

Noting that

$$
\psi_{1} \equiv P_{0}(2 x-1)=1 ; \quad \psi_{2} \equiv P_{1}(2 x-1)=2 x-1,
$$

we find

$$
\begin{gathered}
\left(\psi_{1}, \psi_{1}\right)=\int_{0}^{1} d x=1, \quad\left(\psi_{1}, \psi_{2}\right)=\left(\psi_{2}, \psi_{1}\right)=\int_{0}^{1}(2 x-1) d x=0 \\
\left(\psi_{2}, \psi_{2}\right)=\int_{0}^{1}(2 x-1)^{2} d x=\frac{1}{3}
\end{gathered}
$$

Further,

$$
\begin{aligned}
& \left(K \psi_{1}, \psi_{1}\right)=\int_{0}^{1}\left(\int_{0}^{1} K(x, t) \psi_{1}(t) d t\right) \psi_{1}(x) d x=\int_{0}^{1} \int_{0}^{1} x t d x d t=\frac{1}{4} \\
& \left(K \psi_{1}, \psi_{2}\right)=\int_{0}^{1} \int_{0}^{1} x t(2 x-1) d x d t=\frac{1}{12} \\
& \left(K \psi_{2}, \psi_{2}\right)=\int_{0}^{1} \int_{0}^{1} x t(2 t-1)(2 x-1) d x d t=\frac{1}{36}
\end{aligned}
$$

In this case, system (3) takes the form

$$
\left|\begin{array}{cc}
\frac{1}{4}-\sigma & \frac{1}{12} \\
\frac{1}{12} & \frac{1}{36}-\frac{1}{3} \sigma
\end{array}\right|=0
$$

or

$$
\sigma^{2}-\sigma\left(\frac{1}{12}+\frac{1}{4}\right)=0
$$

From this, $\sigma_{1}=0, \sigma_{2}=\frac{1}{3}$. The largest eigenvalue $\sigma_{2}=\frac{1}{3}$, so the smallest characteristic number $\lambda=\frac{1}{\sigma_{2}}=3$.

## Problems for Independent Solution

Using the Ritz method, find the smallest characteristic numbers of the kernels ( $a=0$, $b=1):$

340. $K(x, t)=x^{2} t^{2} . \quad$ 341. $K(x, t)= \begin{cases}t, & x \geqslant t, \\ x, & x \leqslant t .\end{cases}$
341. $K(x, t)= \begin{cases}\frac{1}{2} x(2-t), & x \leqslant t, \\ \frac{1}{2} t(2-x), & x \geqslant t .\end{cases}$

$2^{\circ}$. Method of Traces. We call the $m$-th trace of the kernel $K(x, t)$ the number

$$
A_{m}=\int_{a}^{b} K_{m}(t, t) d t
$$

where $K_{m}(x, t)$ denotes the $m$-th iterated kernel.

For the smallest characteristic number $\lambda_{1}$, for sufficiently large $m$, the following approximate formula holds:

$$
\left|\lambda_{1}\right| \approx \sqrt{\frac{A_{2 m}}{A_{2 m+2}}}
$$

Formula (4) gives the value of $\left|\lambda_{1}\right|$ with a surplus.

Traces of even order for a symmetric kernel are computed by the formula

$$
A_{2 m}=\int_{a}^{b} \int_{a}^{b} K_{m}^{2}(x, t) d x d t=2 \int_{a}^{b} \int_{a}^{x} K_{m}^{2}(x, t) d t d x
$$