Example 1. Using Fredholm determinants, find the resolvent of the kernel $K(x, t)=x e^{t} ; a=0, b=1$.
Solution. We have $B_{0}(x, t)=x e^{t}$. Further,

$$
\begin{aligned}
& B_{1}(x, t)=\int_{0}^{1}\left|\begin{array}{ll}
x e^{t} & x e^{t_{1}} \\
t_{1} e^{t} & t_{1} e^{t_{1}}
\end{array}\right| d t_{1}=0 \\
& B_{2}(x, t)=\int_{0}^{1} \int_{0}^{1}\left|\begin{array}{lll}
x e^{t} & x e^{t_{1}} & x e^{t_{2}} \\
t_{1} e^{t} & t_{1} e^{t_{1}} & t_{1} e^{t_{2}} \\
t_{2} e^{t_{2}} & t_{2} e^{t_{1}} & t_{2} e^{t_{2}}
\end{array}\right| d t_{1} d t_{2}=0
\end{aligned}
$$

since the determinants under the integral sign are zero. It is obvious that all subsequent $B_{n}(x, t)=0$. We find the coefficients $C_{n}$:

$$
\begin{aligned}
& C_{1}=\int_{0}^{1} K\left(t_{1}, t_{1}\right) d t_{1}=\int_{0}^{0} t_{1} e^{t_{1}} d t_{1}=1 \\
& C_{2}=\int_{0}^{1} \int_{0}^{1}\left|\begin{array}{ll}
t_{1} e^{t_{1}} & t_{1} e^{t_{2}} \\
t_{2} e^{t_{1}} & t_{2} e^{t_{2}}
\end{array}\right| d t_{1} d t_{2}=0
\end{aligned}
$$

It is obvious that all subsequent $C_{n}=0$.

According to formulas (4) and (5) in our case we have

$$
D(x, t ; \lambda)=K(x, t)=x e^{t} ; \quad D(\lambda)=1-\lambda .
$$

Thus,

$$
R(x, t ; \lambda)=\frac{D(x, t ; \lambda)}{D(\lambda)}=\frac{x e^{t}}{1-\lambda}
$$

We apply the obtained result to solve the integral equation

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

According to formula (2)

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

In particular, for $f(x)=e^{-x}$ we get

$$
\varphi(x)=e^{-x}+\frac{\lambda}{1-\lambda} x
$$

## Problems for Independent Solution

73. Show that for the equation

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

the Fredholm determinant

$$
D(\lambda)=1-\frac{\lambda}{3},
$$

and the Fredholm minor

$$
D(x, t ; \lambda)=x t
$$

74. Show that for the equation

$$
\varphi(x)=x+\lambda \int_{0}^{1}\left(x t+t^{2}\right) \varphi(t) d t
$$

we have

$$
D(\lambda)=1-\frac{2}{3} \lambda-\frac{1}{72} \lambda^{2}, \quad D(x, t ; \lambda)=x t+t^{2}+\lambda\left(\frac{1}{2} x t^{2}-\frac{1}{3} x t-\frac{1}{3} t^{2}+\frac{1}{4} t\right)
$$

75. Show that if

$$
K(x, t)=f_{1}(x) f_{2}(t) \quad \int_{a}^{b} f_{1}(x) f_{2}(x) d x=A
$$

then

$$
D(\lambda)=1-\lambda A, \quad D(x, t ; \lambda)=f_{1}(x) f_{2}(t)
$$

and the solution of the corresponding inhomogeneous integral equation with the right-hand side $f(x)$ has the form

$$
\varphi(x)=f(x)+\frac{\lambda f_{1}(x)}{1-\lambda A} \int_{a}^{b} f(t) f_{2}(t) d t
$$

76. Show that if

$$
K(x, t)=f_{1}(x) g_{1}(t)+f_{2}(x) g_{2}(t)
$$

then $D(\lambda)$ will be a polynomial of the second degree in $\lambda$; generally, if

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

then $D(\lambda)$ will be a polynomial of degree $n$ in $\lambda$.

Using Fredholm determinants, find the resolvents of the following kernels:

77. $K(x, t)=2 x-t ; \quad 0 \leqslant x \leqslant 1,0 \leqslant t \leqslant 1$.
78. $K(x, t)=x^{2} t-x t^{2} ; \quad 0 \leqslant x \leqslant 1,0 \leqslant t \leqslant 1$.
79. $K(x, t)=\sin x \cos t ; \quad 0 \leqslant x \leqslant 2 \pi, 0 \leqslant t \leqslant 2 \pi$.
80. $K(x, t)=\sin x-\sin t ; 0 \leqslant x \leqslant 2 \pi, 0 \leqslant t \leqslant 2 \pi$.

Calculation of the coefficients $B_{n}(x, t)$ and $C_{n}$ of the series (4) and (5) by formulas (6) and (7) is practically possible only in very rare cases, but from these formulas the following recurrence relations are obtained:

$$
\begin{gathered}
B_{n}(x, t)=C_{n} K(x, t)-n \int_{a}^{b} K(x, s) B_{n-1}(s, t) d s \\
C_{n}=\int_{a}^{b} B_{n-1}(s, s) d s
\end{gathered}
$$

Knowing that the coefficient $C_{0}=1$ and $B_{0}(x, t)=K(x, t)$, we can find sequentially $C_{1}, B_{1}(x, t), C_{2}, B_{2}(x, t), C_{3}$, and so on.