Example 4. Find the characteristic numbers and eigenfunctions of the homogeneous equation

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

where

$$
K(x, t)= \begin{cases}\cos x \sin t, & 0 \leqslant x \leqslant t, \\ \cos t \sin x, & t \leqslant x \leqslant \pi\end{cases}
$$
Solution. The given equation can be represented as

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

or

$$
\varphi(x)=\lambda \sin x \int_{0}^{x} \varphi(t) \cos t d t+\lambda \cos x \int_{x}^{\pi} \varphi(t) \sin t d t
$$

Differentiating both sides (15), we find

$$
\begin{aligned}
& \varphi^{\prime}(x)=\lambda \cos x \int_{0}^{x} \varphi(t) \cos t d t+\lambda \sin x \cos x \varphi(x)- \\
&-\lambda \sin x \int_{x}^{x} \varphi(t) \sin t d t-\lambda \sin x \cos x \varphi(x)
\end{aligned}
$$

or

$$
\varphi^{\prime}(x)=\lambda \cos x \int_{0}^{z} \varphi(t) \cos t d t-\lambda \sin x \int_{x}^{\pi} \varphi(t) \sin t d t
$$

A second differentiation gives

$$
\varphi^{\prime \prime}(x)=-\lambda \sin x \int_{0}^{x} \varphi(t) \cos t d t+\lambda \cos ^{2} x \varphi(x)
$$

$$
\begin{gathered}
-\lambda \cos x \int_{x}^{\pi} \varphi(t) \sin t d t+\lambda \sin ^{2} x \varphi(x)= \\
=\lambda \varphi(x)-\left[\lambda \sin x \int_{0}^{\pi} \varphi(t) \cos t d t+\lambda \cos x \int_{x}^{\pi} \varphi(t) \sin t d t\right]
\end{gathered}
$$

The expression in square brackets equals $\varphi(x)$, so

$$
\varphi^{\prime \prime}(x)=\lambda \varphi(x)-\varphi(x)
$$

From equalities (15) and (16), we find that

$$
\varphi(\pi)=0, \quad \varphi^{\prime}(0)=0
$$

Thus, the given integral equation reduces to the following boundary value problem:

$$
\begin{gathered}
\varphi^{\prime \prime}(x)-(\lambda-1) \varphi(x)=0 \\
\varphi(\pi)=0, \quad \varphi^{\prime}(0)=0
\end{gathered}
$$

There are three possible cases.

1) $\lambda-1=0$, or $\lambda=1$. Equation (17) becomes $\varphi^{\prime \prime}(x)=0$. Its general solution is $\varphi(x)=C_{1} x+C_{2}$. Using the boundary conditions (18), we get the system for finding the unknowns $C_{1}$ and $C_{2}$

$$
\left\{\begin{array}{r}
C_{1} \pi+C_{2}=0 \\
C_{1}=0
\end{array}\right.
$$

which has the unique solution $C_{1}=0, C_{2}=0$, and thus the integral equation has only the trivial solution

$$
\varphi(x) \equiv 0
$$

2) $\lambda-1>0$, or $\lambda>1$. The general solution of equation (17) is

$$
\varphi(x)=C_{1} \operatorname{ch} \sqrt{\lambda-1} x+C_{2} \operatorname{sh} \sqrt{\lambda-1} x
$$

from which

$$
\varphi^{\prime}(x)=\sqrt{\lambda-1}\left(C_{1} \operatorname{sh} \sqrt{\lambda-1} x+C_{2} \operatorname{ch} \sqrt{\lambda-1} x\right)
$$

The boundary conditions give the system for finding the values of $C_{1}$ and $C_{2}$

$$
\left\{\begin{aligned}
C_{1} \operatorname{ch} \pi \sqrt{\lambda-1}+C_{2} \operatorname{sh} \pi \sqrt{\lambda-1} & =0 \\
C_{2} & =0
\end{aligned}\right.
$$

The system has the unique solution $C_{1}=0, C_{2}=0$. The integral equation has the trivial solution $\varphi(x) \equiv 0$. Thus, for $\lambda>1$, the integral equation has no eigenvalues, and therefore no eigenfunctions.
3) $\lambda-1<0$, or $\lambda<1$. The general solution of equation (17) is

$$
\varphi(x)=C_{1} \cos \sqrt{1-\lambda} x+C_{2} \sin \sqrt{1-\lambda} x
$$

From this, we find that

$$
\varphi^{\prime}(x)=\sqrt{1-\lambda}\left(-C_{1} \sin \sqrt{1-\lambda} x+C_{2} \cos \sqrt{1-\lambda} x\right)
$$

The boundary conditions (18) in this case give the system for finding $C_{1}$ and $C_{2}$

$$
\left\{\begin{array}{r}
C_{1} \cos \pi \sqrt{1-\lambda}+C_{2} \sin \pi \sqrt{1-\lambda}=0 \\
\sqrt{1-\lambda} C_{2}=0
\end{array}\right.
$$

The determinant of this system is

$$
\Delta(\lambda)=\left|\begin{array}{cc}
\cos \pi \sqrt{1-\lambda} & \sin \pi \sqrt{1-\lambda} \\
0 & \sqrt{1-\lambda}
\end{array}\right|
$$

Setting it equal to zero, we get the equation for finding the eigenvalues:

$$
\left|\begin{array}{cc}
\cos \pi \sqrt{1-\lambda} & \sin \pi \sqrt{1-\lambda} \\
0 & \sqrt{1-\lambda}
\end{array}\right|=0
$$

or $\sqrt{1-\lambda} \cos \pi \sqrt{1-\lambda}=0$. By assumption, $\sqrt{1-\lambda} \neq 0$, so $\cos \pi \sqrt{1-\lambda}=0$. From this, we find that $\pi \sqrt{1-\lambda}=\frac{\pi}{2}+\pi n$, where $n$ is any integer. All roots of equation (20) are given by the formula

$$
\lambda_{n}=1-\left(n+\frac{1}{2}\right)^{2}
$$

For the values $\lambda=\lambda_{n}$, the system (19) becomes

$$
\left\{\begin{aligned}
C_{1} \cdot 0 & =0 \\
C_{2} & =0
\end{aligned}\right.
$$

It has an infinite set of non-zero solutions

$$
\left\{\begin{array}{l}
C_{1}=C \\
C_{2}=0
\end{array}\right.
$$

where $C$ is any constant. Therefore, the original integral equation has an infinite set of solutions of the form

$$
\varphi(x)=C \cos \left(n+\frac{1}{2}\right) x
$$

which are the eigenfunctions of this equation.

Thus, the eigenvalues and eigenfunctions of the given integral equation are

$$
\lambda_{n}=1-\left(n+\frac{1}{2}\right)^{2}, \quad \varphi_{n}(x)=\cos \left(n+\frac{1}{2}\right) x
$$

where $n$ is any integer.