Example 2. Find the iterated kernels $K_{1}(x, t)$ and $K_{2}(x, t)$, if $K(x, t)=e^{\min (x, t)}, a=0, b=1$.
Solution. By definition, we have

$$
\min \{x, t\}= \begin{cases}x, & \text { if } \quad 0 \leqslant x \leqslant t \\ t, & \text { if } t \leqslant x \leqslant 1\end{cases}
$$

therefore, the given kernel can be written as

![](https://cdn.mathpix.com/cropped/2024_05_22_d528bb4cd8a42024c50cg-044.jpg?height=100&width=484&top_left_y=1159&top_left_x=369)

This kernel, as it is easy to verify, is symmetric, i.e.,

$$
K(x, t)=K(t, x)
$$

We have $K_{1}(x, t)=K(x, t)$. We find the second iterated kernel:

$$
K_{2}(x, t)=\int_{0}^{1} K(x, s) K_{1}(s, t) d s=\int_{0}^{1} K(x, s) K(s, t) d s
$$

Here,

$$
\begin{aligned}
K(x, s) & =\left\{\begin{array}{lll}
e^{x}, & \text { if } & 0 \leqslant x \leqslant s, \\
e^{s}, & \text { if } & s \leqslant x \leqslant 1,
\end{array}\right. \\
K(s, t) & =\left\{\begin{array}{lll}
e^{s}, & \text { if } & 0 \leqslant s \leqslant t, \\
e^{t}, & \text { if } t \leqslant s \leqslant 1 .
\end{array}\right.
\end{aligned}
$$

Since the kernel $K(x, t)$ is symmetric, it is sufficient to find $K_{2}(x, t)$ only for $x > t$.

We have (Fig. 2)

$$
K_{2}(x, t)=\int_{0}^{t} K(x, s) K(s, t) d s+\int_{t}^{x} K(x, s) K(s, t) d s+\int_{x}^{1} K(x, s) K(s, t) d s
$$

In the interval $(0, t)$, we have $s < x > t$, therefore

$$
\int_{x}^{1} K(x, s) K(s, t) d s=\int_{x}^{1} e^{x} e^{t} d s=(1-x) e^{x+t}
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_d528bb4cd8a42024c50cg-045.jpg?height=276&width=524&top_left_y=1001&top_left_x=106)

Fig. 2
By summing the found integrals, we get

$$
K_{2}(x, t)=(2-x) e^{x+t}-\frac{1+e^{2 t}}{2} \quad(x > t)
$$

The expression for $K_{2}(x, t)$ when $x < t$ is:

$$
K_{2}(x, t)=(2-t) e^{x+t}-\frac{1+e^{2 x}}{2} \quad(x < t)
$$

Thus, the second iterated kernel is given by

$$
K_{2}(x, t)=\left\{\begin{array}{lc}
(2-t) e^{x+t}-\frac{1+e^{2 x}}{2}, & \text { if } 0 \leqslant x \leqslant t \\
(2-x) e^{x+t}-\frac{1+e^{2 t}}{2}, & \text { if } t \leqslant x \leqslant 1
\end{array}\right.
$$

Remark. If the kernel $K(x, t)$, defined in the square $a \leqslant x \leqslant b, a \leqslant t \leqslant b$ by different analytical expressions, is not symmetric, then the case $x < t$ should be considered separately. For $x < t$, we will have (Fig. 3)

$$
K_{2}(x, t)=\int_{a}^{b} K(x, s) K(s, t) d s=\int_{a}^{x}+\int_{x}^{t}+\int_{t}^{b}
$$