Notes on Whittaker & Watson, Chapter XI, part 2

Whittaker & Watson, A Course of Modern Analysis. Chapter XI, Integral Equations.

11.21 Verification of the solution to Fredholm’s equation of the second kind from 11.2.

By Hadamard’s inequality (11.11) series expansion of $D(\lambda)$ converges absolutely for all $\lambda$, and $D(x,y; \lambda)-\lambda K(x,y)$ is uniformly convergent function of $x,y$.

Factorizing $K(x, y)$ from $D(x,y; \lambda)$ $$ \begin{align*} D(x,y;\lambda) = \lambda D(\lambda)K(x,y) & - \lambda^2 \int_a^b\begin{vmatrix} 0 & K(x,\xi_1)\newline K(\xi_1,y) & K(\xi_1,\xi_1) \end{vmatrix} d\xi_1 \newline & + \frac{\lambda^3}{2!} \int_a^b\int_a^b\begin{vmatrix} 0 & K(x,\xi_1) & K(x,\xi_2)\newline K(\xi_1,y) & K(\xi_1,\xi_1) & K(\xi_1,\xi_2)\newline K(\xi_2,y) & K(\xi_2,\xi_1) & K(\xi_2,\xi_2) \end{vmatrix}d\xi_1d\xi_2 - \cdots. \end{align*} $$ The integral of the determinant in each term $$ \begin{align*} & \int_a^b\int_a^b\cdots\int_a^b\begin{vmatrix} 0 & K(x,\xi_1) & \cdots & K(x,\xi_n)\newline K(\xi_1,y) & K(\xi_1,\xi_1) & \cdots & K(\xi_1,\xi_n)\newline \cdots & \cdots & \cdots & \cdots \newline K(\xi_n,y) & K(\xi_n,\xi_1) & \cdots & K(\xi_n,\xi_n) \end{vmatrix}d\xi_1d\xi_2\cdots d\xi_n \newline =& -n \int_a^b\int_a^b\cdots\int_a^bK(\xi,y) \begin{vmatrix} K(x,\xi) & K(x,\xi_1) & \cdots & K(x,\xi_{n-1})\newline K(\xi_1,\xi) & K(\xi_1,\xi_1) & \cdots & K(\xi_1,\xi_{n-1})\newline \cdots & \cdots & \cdots \newline K(\xi_{n-1},\xi) & K(\xi_{n-1},\xi_1) & \cdots & K(\xi_{n-1},\xi_{n-1}) \end{vmatrix} d\xi d\xi_1\cdots d\xi_{n-1} \end{align*} $$ results in $$ D(x,y;\lambda) = \lambda D(\lambda)K(x,y) + \lambda\int_a^bK(\xi,y)D(x,\xi;\lambda)d\xi $$ Integrating the original equation $\phi(\xi)=f(\xi)+\lambda\int_a^bK(\xi,y)\phi(y)dy$ with factor $D(x,\xi;\lambda)d\xi$ on both sides, we have $$ \int_a^bf(\xi)D(x,\xi;\lambda)d\xi = D(\lambda)(\phi(x)-f(x)) $$ which verifies the solution $\phi(x) = f(x)+\int_a^b\frac{D(x,\xi;\lambda)}{D(\lambda)}f(\xi)d\xi$, as long as $D(\lambda)\neq 0$.

11.22 Volterra’s reciprocal functions. Two kernels $K(x, y)$ and $k(x,y;\lambda)$ are reciprocal if $$ K(x, y) + k(x,y;\lambda)=\lambda\int_a^bk(x,\xi;\lambda)K(\xi,y)d\xi. $$ Thus in Fredholm’s equation, $K(x,y)$ and $k(x,y;\lambda)=-\frac{D(x,y; \lambda)}{\lambda D(\lambda)}$ are reciprocal.

Integrating Fredholm’s equation by kernel $k(\cdot,\cdot;\lambda)$ $$ \begin{align*} \int_a^bk(x,\xi;\lambda)\phi(\xi)d\xi &= \int_a^bk(x,\xi;\lambda)f(\xi)d\xi+\lambda\int_a^b\int_a^bk(x,\xi;\lambda)K(\xi,\xi_1)\phi(\xi_1)d\xi_1 d\xi\newline &=\int_a^bk(x,\xi;\lambda)f(\xi)d\xi+\int_a^b\left[K(x,\xi_1)+k(x,\xi_1;\lambda)\right]\phi(\xi_1)d\xi_1, \end{align*} $$ in which the LHS cancels the same term on the RHS $$ \lambda\int_a^bk(x,\xi;\lambda)f(\xi)d\xi = -\lambda\int_a^bK(x,\xi)\phi(\xi)d\xi=f(x)-\phi(x) $$ we have the the ‘dual’ equation $f(x)=\phi(x)+\lambda\int_a^bk(x,\xi;\lambda)f(\xi)d\xi$.

11.23 Homogeneous integral equations. When $D(\lambda)\neq 0$, the homogeneous equation $$ \phi(x) = \int_a^bK(x,\xi)\phi(\xi)d\xi $$ has the only continuous solution $\phi(x)=0$.

The roots of $D(\lambda)=0$ are called the characteristic numbers of the kernel. Expanding $D(\lambda)$ at the root $\lambda=\lambda_0$ with multiplicity $m\ge 1$, we have $$ D(\lambda) = c_m(\lambda-\lambda_0)^m+c_{m+1}(\lambda-\lambda_0)^{m+1}+\cdots \qquad (c_m\neq0) $$ and $$ D(x,y;\lambda)=\frac{g_l(x,y)}{l!}(\lambda-\lambda_0)^l + \frac{g_{l+1}(x,y)}{(l+1)!}(\lambda-\lambda_0)^{l+1}+\cdots\qquad(l\ge 0, g_l\not\equiv0). $$ Substituting the Taylor series for $D(\lambda)$ and $D(x,y;\lambda)$ in the result from 11.21, we have $$ g_l(x,y) =\lambda_0\int_a^bK(x,\xi)g_l(\xi, y)d\xi $$ showing that the linear combinations of $g_l(\cdot, y)$ satisfy the homogeneous integral equation.