Example 4. Find the resolvent for the kernel

$$
K(x, t)=x t+x^{2} t^{2}, \quad a=-1, \quad b=1
$$
Solution. As shown above, the kernels $M(x, t)=x t$ and $N(x, t)=x^{2} t^{2}$ are orthogonal on $[-1,1]$ (see p. 41). Therefore, the resolvent of the kernel $K(x, t)$ is equal to the sum of the resolvents of the kernels $M(x, t)$ and $N(x, t)$. Using the results of problems 104 and 105, we find

$$
R_{K}(x, t ; \lambda)=R_{M}(x, t ; \lambda)+R_{N}(x, t ; \lambda)=\frac{3 x t}{3-2 \lambda}+\frac{5 x^{2} t^{2}}{5-2 \lambda}
$$

where $|\lambda|<\frac{3}{2}$.

## Problems for Independent Solution

Find the resolvents for the kernels:

106. $K(x, t)=\sin x \cos t+\cos 2 x \sin 2 t ; \quad a=0, b=2 \pi$.
107. $K(x, t)=1+(2 x-1)(2 t-1) ; \quad a=0, b=1$.

The specified property can be extended to any finite number of kernels.

If the kernels $M^{(1)}(x, t), M^{(2)}(x, t), \ldots, M^{(n)}(x, t)$ are pairwise orthogonal, then the resolvent corresponding to their sum

$$
K(x, t)=\sum_{m=1}^{n} M^{(m)}(x, t)
$$

is equal to the sum of the resolvents corresponding to each of the terms.

We will call the $n$-th trace of the kernel $K(x, t)$ the quantity

$$
A_{n}=\int_{a}^{b} K_{n}(x, x) d x \quad(n=1,2 \ldots)
$$

where $K_{n}(x, t)$ is the $n$-th iterated kernel for the kernel $K(x, t)$.

The following formula holds for the Fredholm determinant $D(\lambda)$:

$$
\frac{D^{\prime}(\lambda)}{D(\lambda)}=-\sum_{n=1}^{\infty} A_{n} \lambda^{n-1}
$$

The radius of convergence of the power series (13) is equal to the smallest of the moduli of the characteristic numbers.

## Problems for Independent Solution

108. Show that for the Volterra equation

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

the Fredholm determinant $D(\lambda)=e^{-A_{1} \lambda}$, and, consequently, the resolvent for the Volterra equation is an entire analytic function of $\lambda$.

109. Let $R(x, t ; \lambda)$ be the resolvent of some kernel $K(x, t)$.

Show that the resolvent of the equation

$$
\varphi(x)-\mu \int_{a}^{b} R(x, t ; \lambda) \varphi(t) d t^{\prime}=f(x)
$$

is equal to $R(x, t ; \lambda+\mu)$.

110. Let

$$
\int_{a}^{b} \int_{a}^{b} K^{2}(x, t) d x d t=B^{2}, \quad \int_{a}^{b} \int_{a}^{b} K_{n}^{2}(x, t) d x d t=B_{n}^{2}
$$

where $K_{n}(x, t)$ is the $n$-th iterated kernel for the kernel $K(x, t)$. Prove that if $B_{2}=B^{2}$, then for any $n$ it will be $B_{n}=B^{n}$.

## § 9. Integral Equations with Degenerate Kernel

The kernel $K(x, t)$ of the Fredholm integral equation of the second kind is called degenerate if it is a sum of a finite number of products of functions of $x$ only by functions of $t$ only, i.e., if it has the form

$$
K(x, t)=\sum_{k=1}^{n} a_{k}(x) b_{k}(t)
$$

the functions $a_{k}(x)$ and $b_{k}(t)(k=1,2, \ldots, n)$ will be considered continuous in the main square $a \leqslant x, t \leqslant b$ and linearly independent of each other. The integral equation with a degenerate kernel (1)

$$
\varphi(x)-\lambda \int_{a}^{b}\left[\sum_{k=1}^{n} a_{k}(x) b_{k}(t)\right] \varphi(t) d t=f(x)
$$

is solved as follows

Rewrite (2) in the form

$$
\varphi(x)=f(x)+\lambda \sum_{k=1}^{n} a_{k}(x) \int_{a}^{b} b_{k}(t) \varphi(t) d t
$$

and introduce the notation

$$
\int_{a}^{b} b_{k}(t) \varphi(t) d t=C_{k} \quad(k=1,2, \ldots, n)
$$

Then (3) will take the form

$$
\varphi(x)=f(x)+\lambda \sum_{k=1}^{n} C_{k} a_{k}(x)
$$

where $C_{k}$ is an unknown constant (since the function $\varphi(x)$ is unknown).

Thus, solving the integral equation with a degenerate kernel reduces to finding the constants $C_{k}(k=1,2, \ldots, n)$. Substituting the expression (5) into the integral equation (2), after some calculations, we obtain

$$
\sum_{m=1}^{n}\left\{C_{m}-\int_{a}^{b} b_{m}(t)\left[f(t)+\lambda \sum_{k=1}^{n} C_{k} a_{k}(t)\right] d t\right\} a_{m}(x)=0
$$

Due to the linear independence of the functions $a_{m}(x)(m=1,2, \ldots, n)$, it follows that

$$
C_{m}-\int_{a}^{b} b_{m}(t)\left[f(t)+\lambda \sum_{k=1}^{n} C_{k} a_{k}(t)\right] d t=0
$$

or

$$
C_{m}-\lambda \sum_{k=1}^{n} C_{k} \int_{a}^{b} a_{k}(t) b_{m}(t) d t=\int_{a}^{b} b_{m}(t) f(t) d t \quad(m=1,2, \ldots, n)
$$

Introducing for brevity the notation

$$
a_{k m}=\int_{a}^{b} a_{k}(t) b_{m}(t) d t, \quad f_{m}=\int_{a}^{b} b_{m}(t) f(t) d t
$$

we get that

$$
C_{m}-\lambda \sum_{k=1}^{n} a_{k m} C_{k}=f_{m} \quad(m=1,2, \ldots, n)
$$

or in expanded form:

$$
\left\{\begin{array}{c}
\left(1-\lambda a_{11}\right) C_{1}-\lambda a_{12} C_{2}-\ldots-\lambda a_{1 n} C_{n}=f_{1} \\
-\lambda a_{21} C_{1}+\left(1-\lambda a_{22}\right) C_{2}-\ldots-\lambda a_{2 n} C_{n}=f_{2} \\
\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \\
-\lambda a_{n 1} C_{1}-\lambda a_{n 2} C_{2}-\ldots+\left(1-\lambda a_{n n}\right) C_{n}=f_{n}
\end{array}\right.
$$

To find the unknowns $C_{k}$, we have a linear system of $n$ algebraic equations with $n$ unknowns. The determinant of this system is equal to

$$
\Delta(\lambda)=\left|\begin{array}{rrrr}
1-\lambda a_{11} & -\lambda a_{12} & \ldots & -\lambda a_{1 n} \\
-\lambda a_{21} & 1-\lambda a_{22} & \ldots & -\lambda a_{2 n} \\
\cdots \cdots \cdots \cdots & \ldots \ldots \ldots \ldots & \ldots & \ldots \\
-\lambda a_{n 1} & -\lambda a_{n 2} & \ldots & 1-\lambda a_{n n}
\end{array}\right|
$$

If $\Delta(\lambda) \neq 0$, then the system (6) has a unique solution $C_{1}, C_{2}, \ldots, C_{n}$, obtained by Cramer's formulas

$$
\begin{aligned}
& C_{k}