Example 2. Using formulas (8) and (9), find the resolvent of the kernel $K(x, t)=x-2 t$, where $0 \leqslant x \leqslant 1,0 \leqslant t \leqslant 1$.
Solution. We have $C_{0}=1, B_{0}(x, t)=x-2 t$. Using formula (9), we find

$$
C_{1}=\int_{0}^{1}(-s) d s=-\frac{1}{2}
$$

By formula (8), we get

$$
B_{1}(x, t)=-\frac{x-2 t}{2}-\int_{0}^{1}(x-2 s)(s-2 t) d s=-x-t+2 x t+\frac{2}{3}
$$

Further, we will have

$$
\begin{gathered}
C_{2}=\int_{0}^{1}\left(-2 s+2 s^{2}+\frac{2}{3}\right) d s=\frac{1}{3} \\
B_{2}(x, t)=\frac{x-2 t}{3}-2 \int_{0}^{1}(x-2 s)\left(-s-t+2 s t+\frac{2}{3}\right) d s=0 \\
C_{3}=C_{4}=\ldots=0, \quad B_{3}(x, t)=B_{4}(x, t)=\ldots=0
\end{gathered}
$$

Therefore,

$$
D(\lambda)=1+\frac{\lambda}{2}+\frac{\lambda^{2}}{6} ; \quad D(x, t ; \lambda)=x-2 t+\left(x+t-2 x t-\frac{2}{3}\right) \lambda .
$$

The resolvent of the given kernel will be

$$
R(x, t ; \lambda)=\frac{x-2 t+\left(x+t-2 x t-\frac{2}{3}\right) \lambda}{1+\frac{\lambda}{2}+\frac{\lambda^{2}}{6}}
$$

## Problems for Independent Solution

Using the recurrence relations (8) and (9), find the resolvents of the following kernels:

81. $K(x, t)=x+t+1 ; \quad-1 \leqslant x \leqslant 1, \quad-1 \leqslant t \leqslant 1$.
82. $K(x, t)=1+3 x t ; \quad-0 \leqslant x \leqslant 1, \quad 0 \leqslant t \leqslant 1$.
83. $K(x, t)=4 x t-x^{2} ; \quad-0 \leqslant x \leqslant 1, \quad 0 \leqslant t \leqslant 1$.
84. $K(x, t)=e^{x-t} ; \quad-0 \leqslant x \leqslant 1, \quad 0 \leqslant t \leqslant 1$.
85. $K(x, t)=\sin (x+t) ; \quad-0 \leqslant x \leqslant 2 \pi, \quad 0 \leqslant t \leqslant 2 \pi$.
86. $K(x, t)=x-\operatorname{sh} t ; \quad-1 \leqslant x \leqslant 1, \quad-1 \leqslant t \leqslant 1$.

Using the resolvent, solve the following integral equations:

87. $\varphi(x)-\lambda \int_{0}^{2 \pi} \sin (x+t) \varphi(t) d t=1$.
88. $\varphi(x)-\lambda \int_{0}^{1}(2 x-t) \varphi(t) d t=\frac{x}{6}$.
89. $\varphi(x)-\int_{0}^{2 \pi} \sin x \cos t \varphi(t) d t=\cos 2 x . \quad 90 . \varphi(x)+\int_{0}^{1} e^{x-t} \varphi(t) d t=e^{x}$.
90. $\varphi(x)-\lambda \int_{0}^{1}\left(4 x t-x^{2}\right) \varphi(t) d t=x$.

## 8 8. Iterated Kernels. Construction of the Resolvent Using Iterated Kernels

Let us have an integral equation of Fredholm

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

As in the case of Volterra equations, the integral equation (1) can be solved by the method of successive approximations. For this, we assume

$$
\varphi(x)=f(x)+\sum_{n=1}^{\infty} \psi_{n}(x) \lambda^{n}
$$

where $\psi_{n}(x)$ are defined by the formulas

$$
\begin{aligned}
& \psi_{1}(x)=\int_{a}^{b} K(x, t) f(t) d t, \\
& \psi_{2}(x)=\int_{a}^{b} K(x, t) \psi_{1}(t) d t=\int_{a}^{b} K_{2}(x, t) f(t) d t_{v_{1}}^{b} \\
& \psi_{3}(x)=\int_{a}^{b} K(x, t) \psi_{2}(t) d t=\int_{a} K_{3}(x, t) f(t) d t \quad \text { and so on. }
\end{aligned}
$$

Here

$$
K_{2}(x, t)=\int_{a}^{b} K(x, z) K_{1}(z, t) d z, \quad K_{3}(x, t)=\int_{a}^{b} K(x ; z) K_{2}(z, t) d z
$$

and generally

$$
K_{n}(x, t)=\int_{a}^{b} K(x, z) K_{n-1}(z, t) d z
$$

$n=2,3, \ldots$, where $K_{1}(x, t) \equiv K(x, t)$. The functions $K_{n}(x, t)$, defined by formulas (3), are called iterated kernels. For them, the relation holds

$$
K_{n}(x, t)=\int_{a}^{b} K_{m}(x, s) K_{n-m}(s, t) d s
$$

where $m$ is any natural number less than $n$.

The resolvent of the integral equation (1) is defined through the iterated kernels by the formula

$$
R(x, t ; \lambda)=\sum_{n=1}^{\infty} K_{n}(x, t) \lambda^{n-1}
$$

where the series on the right-hand side is called the Neumann series of the kernel $K(x, t)$. It converges for

$$
|\lambda|\frac{1}{B}$.

Consider an example:

$$
\varphi(x)-\lambda \int_{0}^{1} \varphi(t) d t=1
$$

Here $K(x, t) \equiv 1$, and therefore,

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

Thus, condition (6) gives that the series (5) converges for $|\lambda|1$ the equation (8) is solvable. Indeed, if $\lambda \neq 1$, then the function $\varphi(x)=\frac{1}{1-\lambda}$ is a solution of the given equation, which is easily verified by direct substitution.

For some Fredholm equations, the Neumann series (5) for the resolvent converges for any values of $\lambda$. Let's show this.

Let us have two kernels: $K(x, t)$ and $L(x, t)$. We will call these kernels orthogonal if the following two conditions are satisfied:

$$
\int_{a}^{b} K(x, z) L(z, t) d z=0, \quad \int_{a}^{b} L(x, z) K(z, t) d z=0
$$

for any admissible values of $x$ and $t$.

For example, the kernels $K(x, t)=x t$ and $L(x, t)=x^{2} t^{2}$ are orthogonal on $[-1,1]$.

Indeed,

$$
\int_{-1}^{1}(x z)\left(z^{2} t^{2}\right) d z=x t^{2} \int_{-1}^{1} z^{3} d z=0, \quad \int_{-1}^{1}\left(x^{2} z^{2}\right)(z t) d z=x^{2} t \int_{-1}^{1} z^{3} d z=0
$$

There are also kernels that are orthogonal to themselves. For such kernels, $K_{2}(x, t) \equiv 0$, where $K_{2}(x, t)$ is the second iterated kernel. In this case, obviously, all subsequent iterated kernels are also zero, and the resolvent coincides with the kernel $K(x, t)$.