Example 3. Using the Kellogg method, calculate the smallest characteristic number of the kernel $K(x, t)=x^{2} t^{2}, 0 \leqslant x, t \leqslant 1$.
Solution. Let $\omega(x)=x$. Then

$$
\begin{aligned}
& \omega_{1}(x)=\int_{0}^{1} x^{2} t^{2} t d t=\frac{x^{2}}{4} \\
& \omega_{2}(x)=\int_{0}^{1} x^{2} t^{4} \frac{1}{4} d t=\frac{1}{4} x^{2} \cdot \frac{1}{5} \\
& \omega_{3}(x)=\int_{0}^{1} \frac{1}{4 \cdot 5} x^{2} t^{4} d t=\frac{1}{4 \cdot 5^{2}} x^{2} \\
& \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \\
& \omega_{n}(x)=\frac{1}{4 \cdot 5^{n-1}} x^{2}
\end{aligned}
$$

Next,

$$
\left\|\omega_{n}(x)\right\|=\frac{1}{4 \cdot 5^{n-1}} \sqrt{\int_{0}^{1} x^{4} d x}=\frac{1}{4 \cdot 5^{n-1}} \cdot \frac{1}{\sqrt{5}}
$$

Thus, according to (7),

$$
\lambda_{1} \approx \frac{\frac{1}{4} \cdot \frac{1}{5^{n-2}} \cdot \frac{1}{\sqrt{5}}}{\frac{1}{4} \cdot \frac{1}{5^{n-1}} \cdot \frac{1}{\sqrt{5}}}=5
$$

## Problems for Independent Solution

Using the Kellogg method, find the smallest characteristic numbers of the following kernels:

347. $K(x, t)=x t$;

$0 \leqslant x, t \leqslant 1$.

348. $K(x, t)=\sin x \sin t$;

$-\pi \leqslant x, t \leqslant \pi$.

349. $K(x, t)=\left\{\begin{array}{ll}t, & x \geqslant t ; \\ x, & x \leqslant t ;\end{array} \quad 0 \leqslant x, t \leqslant 1\right.$.
350. $K(x, t)=\left\{\begin{array}{l}\frac{1}{2} x(2-t), \quad x \leqslant t, \\ \frac{1}{2} t(2-x), \quad x \geqslant t,\end{array} \quad 0 \leqslant x, t \leqslant 1\right.$.

Stability of a Compressed Rod (Longitudinal Bending of a Rod)

The equation of the deflected elastic line of the rod is

$$
\frac{d}{d x}\left(E I \frac{d y}{d x}\right)=M
$$

where $\boldsymbol{M}$ is the bending moment, $I$ is the moment of inertia of the cross-section of the rod with abscissa $x$, and $E$ is the Young's modulus.

Consider the case when the rod is compressed by forces applied to its ends. The magnitude of each of these forces is denoted by $P$. Then $M=-\boldsymbol{P} \cdot \boldsymbol{y}$ and equation (9) becomes

$$
\frac{d}{d x}\left(E I \frac{d y}{d x}\right)+P \cdot y=0
$$

The ends of the rod do not move in the direction perpendicular to the rod, so

$$
y(0)=y(l)=0,
$$

where $l$ is the length of the rod. Dividing both sides of equation (10) by $E$ and setting $\frac{\boldsymbol{P}}{\boldsymbol{E}}=\lambda$, we get

$$
\frac{d}{d x}\left(r \frac{d y}{d x}\right)+\lambda y=0
$$

Let $G(x, t)$ be the Green's function of the differential equation (12) with boundary conditions (11). Then the problem (12)-(11) is equivalent to the problem of solving the homogeneous Fredholm integral equation of the second kind

$$
y(x)=\lambda \int_{0}^{l} G(x, t) y(t) d t
$$

The kernel $G(x, t)$ of equation (13) is symmetric as the Green's function of a self-adjoint boundary value problem. Thus, the deflection $y(x)$ of the compressed rod satisfies the integral equation (13). For an arbitrarily chosen force $P$, the number $\lambda=\frac{P}{E}$ will not be a characteristic number, and the solution of equation (13) will be the function $y(x) \equiv 0$. In other words, an arbitrarily chosen compressive force will leave the rod straight. In the case when
$P=\lambda_{n} E$, where $\lambda_{n}$ is a characteristic number of the kernel $G(x, t)$, equation (13) will have a non-zero solution, which corresponds to the bending of the rod, i.e., the loss of its stability.

The smallest force $P$ at which the rod loses stability is called the critical force. It is obviously equal to $P_{\text {cr }}=\lambda_{1} E$, where $\lambda_{1}$ is the smallest characteristic number of equation (13). For practical purposes, a sufficiently good approximate formula is

$$
\lambda_{1} \approx \frac{1}{\sqrt{\int_{0}^{1} \int_{0}^{1} G^{2}(x, t) d x d t}}
$$

which gives $\lambda_{1}$ with a deficiency.