Example. Solve the equation by the Bubnov-Galerkin method

$$
\varphi(x)=x+\int_{-1}^{1} x t \varphi(t) d t
$$
Solution. As a complete system of functions on $[-1,1]$, we choose the system of Legendre polynomials $P_{n}(x)(n=0,1,2, \ldots)$. We will seek the approximate solution $\varphi_{n}(x)$ of equation (4) in the form

$$
\varphi_{3}(x)=a_{1} \cdot 1+a_{2} x+a_{3} \frac{3 x^{2}-1}{2}
$$

Substituting $\varphi_{3}(x)$ for $\varphi(x)$ in equation (4), we will have

$$
a_{1}+a_{2} x+a_{3} \frac{3 x^{2}-1}{2}=x+\int_{-1}^{1} x t\left(a_{1}+a_{2} t+a_{3} \frac{3 t^{2}-1}{2}\right) d t
$$

or

$$
a_{1}+a_{2} x+a_{3} \frac{3 x^{2}-1}{2}=x+x \frac{2}{3} a_{2}
$$

Multiplying both sides of (5) successively by $1, x, \frac{3 x^{2}-1}{2}$ and integrating over $x$ from -1 to 1, we find

$$
\begin{aligned}
2 a_{1} & =0 \\
\frac{2}{3} a_{2} & =\frac{2}{3}+\frac{4}{9} a_{2} \\
\frac{2}{5} a_{3} & =0
\end{aligned}
$$

Hence, $a_{1}=0, a_{2}=3, a_{3}=0$, and thus, $\varphi_{3}(x)=3 x$. It is not difficult to verify that this is the exact solution of equation (4).

## Problems for Independent Solution

Using the Bubnov-Galerkin method, solve the following integral equations:

337. $\varphi(x)=1+\int_{-1}^{1}\left(x t+x^{2}\right) \varphi(t) d t . \quad$ 338. $\varphi(x)=1+\frac{4}{3} x+\int_{-1}^{1}\left(x t^{2}-x\right) \varphi(t) d t$.
338. $\varphi(x)=1-x\left(e^{x}-e^{-x}\right)+\int_{-1}^{1} x^{2} e^{x t} \varphi(t) d t$.

Remark. For degenerate kernels, the Bubnov-Galerkin method gives the exact solution, and for the general case, it is equivalent to replacing the kernel $K(x, t)$ with a degenerate $L(x, t)$.

## § 27. Approximate Methods for Finding Characteristic Numbers and Eigenfunctions of Symmetric Kernels

$1^{\circ}$. Ritz Method. Suppose we have the integral equation

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

with a symmetric kernel $K(x, t) \equiv K(t, x)$.

Choose a sequence of functions $\left\{\psi_{n}(x)\right\}, \psi_{n}(x) \in L_{2}(a, b)$, such that the system $\left\{\psi_{n}(x)\right\}$ is complete in $L_{2}(a, b)$ and for any $n$ the functions $\psi_{1}(x), \psi_{2}(x), \ldots, \psi_{n}(x)$ are linearly independent on $[a, b]$. Assume

$$
\varphi_{n}(x)=\sum_{k=1}^{n} a_{k} \psi_{k}(x)
$$

where the coefficients $a_{k}$ are subject to the condition $\left\|\varphi_{n}\right\|=1$, and under this condition, we seek the stationary values of the quadratic form

$$
\left(K \varphi_{n}, \varphi_{n}\right)
$$

We arrive at a homogeneous system of equations for the coefficients $a_{k}$ ( $\sigma-$ Lagrange multiplier):

$$
\sum_{k=1}^{n}\left\{\left(K \psi_{j}, \psi_{k}\right)-\sigma\left(\psi_{j}, \psi_{k}\right)\right\} a_{k}=0 \quad(j=1,2, \ldots, n)
$$

For a non-zero solution to exist, the determinant of the system (2) must be zero:

$$
\left|\begin{array}{ccc}
\left(K \psi_{1}, \psi_{1}\right)-\sigma\left(\psi_{1}, \psi_{1}\right) & \ldots & \left(K \psi_{1}, \psi_{n}\right)-\sigma\left(\psi_{1}, \psi_{n}\right) \\
\left(K \psi_{2}, \psi_{1}\right)-\sigma\left(\psi_{2}, \psi_{1}\right) & \ldots & \left(K \psi_{2}, \psi_{n}\right)-\sigma\left(\psi_{2}, \psi_{n}\right) \\
\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \\
\left(K \psi_{n}, \psi_{1}\right)-\sigma\left(\psi_{n}, \psi_{1}\right) & \ldots & \left(K \psi_{n}, \psi_{n}\right)-\sigma\left(\psi_{n}, \psi_{n}\right)
\end{array}\right|=0
$$

The roots of equation (3) give approximate values of the eigenvalues of the kernel $K(x, t)$. The largest root of equation (3) gives an approximate value of the largest eigenvalue with a deficiency. Finding $\sigma$ from (3) and substituting it into (2), we seek a non-zero solution $a_{k}$ ( $k=1,2, \ldots, n$ ) of the system (2). Substituting the found values of $a_{k}$ into (1), we obtain an approximate expression for the eigenfunction corresponding to the found eigenvalue.