Notes on Whittaker & Watson, Chapter XI, part 1

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

11.1 Definition of Integral Equations.

Examples: solving $\phi$ for $f(x)=\int e^{xt}\phi(t)dt$ (Laplace), $f(x)=\int_{-\infty}^{\infty} \cos{(xt)}\phi(t)dt$ (Fourier).

11.11 Hadamard’s inequality. For square matrix $A = (a_1, a_2, \cdots, a_n)$ with columns normalised $\|a_i\|_2=1, i=1,2,\cdots,n$, the determinant $-1\leq \det(A)\leq 1$.

Collorary. For $A=(a_{ij})$, with each entry $|a_{ij}|<M$, the column vectors $\|a_i\|_2\leq n^{\frac{1}{2}}M$, therefore $|\det(A)|\leq n^{\frac{1}{2}n}M^n$.

11.2 Fredholm’s equation of the second kind is defined as $$ \phi(x)=f(x)+\lambda\int_a^bK(x,\xi)\phi(\xi)d\xi $$ in which $\phi$ is the unknown, $f$ is given, $K(x, \xi)$ is the kernel (the term nucleus was used in Whittaker & Watson) such that $K(\cdot, \cdot)$ is continuous over $[a,b]\times[a,b]$ and $K(x,y)=0$ when $y>x$.

Fredholm’s method first solves the discretized linear system $$ {\boldsymbol\phi} = \mathbf{f} + \lambda\delta\mathbf{K}{\boldsymbol\phi} $$ where ${\boldsymbol\phi}=(\phi(x_0=a), \phi(x_1), \cdots, \phi(x_n=b))^T$ and $\mathbf{K}_{pq}=K(x_p,x_q)$ with $x_q-x_{q-1}=\delta=\frac{b-a}{n}$, then passes the discrete solution ${\boldsymbol\phi}=(\mathbf{I}-\lambda\delta\mathbf{K})^{-1}\mathbf{f}$ to the continuous limit $\delta\to 0$, $n\to\infty$.

The determinant $D_n(\lambda)=\det(\mathbf{I}-\lambda\delta\mathbf{K})$ can be expanded in the power series of $\lambda$ $$ \begin{align*} D_n(\lambda)=&\begin{vmatrix}1- \lambda\delta K(x_1,x_1) & -\lambda\delta K(x_1,x_2) & \cdots & -\lambda\delta K(x_1,x_n)\newline -\lambda\delta K(x_2,x_1) & 1-\lambda\delta K(x_2,x_2) & \cdots & -\lambda\delta K(x_1,x_n)\newline \cdots & \cdots & \cdots & \cdots \newline -\lambda\delta K(x_n,x_1) & -\lambda\delta K(x_n,x_2) & \cdots & 1-\lambda\delta K(x_n,x_n) \end{vmatrix} \newline =& 1- \lambda\sum_{p=1}^n\delta K(x_p,x_p) + \frac{\lambda^2}{2!}\sum_{p,q=1}^n\delta^2\begin{vmatrix} K(x_p,x_p) & K(x_p,x_q)\newline K(x_q,x_p) & K(x_q,x_q) \end{vmatrix} \newline & +\frac{\lambda^3}{3!}\sum_{p,q,r=1}^n\delta^2\begin{vmatrix} K(x_p,x_p) & K(x_p,x_q) & K(x_p,x_r)\newline K(x_q,x_p) & K(x_q,x_q) & K(x_q,x_r)\newline K(x_r,x_p) & K(x_r,x_q) & K(x_r,x_r) \end{vmatrix} - \cdots, \end{align*} $$ with the expansion of cofactor $D_n(x_\mu, x_\nu)$ when $\mu\neq\nu$ $$ D_n(x_\mu, x_\nu) = \lambda\delta\left(K(x_\mu,x_\nu) -\lambda\delta\sum_{p=1}^n\begin{vmatrix} K(x_\mu,x_\nu) & K(x_\mu,x_p)\newline K(x_p,x_\nu) & K(x_p,x_p) \end{vmatrix} \newline +\frac{\lambda^2\delta^2}{2!}\sum_{p,q=1}^n\begin{vmatrix} K(x_\mu,x_\nu) & K(x_\mu,x_p) & K(x_\mu,x_q)\newline K(x_p,x_\nu) & K(x_p,x_p) & K(x_p,x_q)\newline K(x_q,x_\nu) & K(x_q,x_p) & K(x_q,x_q) \end{vmatrix} - \cdots\right). $$ In continuous limit $\delta\sum_{1}^{n}\to\int_a^bd\xi$ $$ D_n(\lambda)\to D(\lambda)=1-\lambda\int_a^bK(\xi_1,\xi_1)d\xi_1+\frac{\lambda^2}{2!}\lambda\int_a^b\int_a^b\begin{vmatrix} K(\xi_1,\xi_1) & K(\xi_1,\xi_2)\newline K(\xi_2,\xi_1) & K(\xi_2,\xi_2) \end{vmatrix}d\xi_1d\xi_2 - \cdots, $$ and $$ \delta^{-1}D_n(x_\mu,x_\nu)\to D(x_\mu,x_\nu;\lambda) =\lambda K(x_\mu,x_\nu) -\lambda^2\int_a^b\begin{vmatrix} K(x_\mu,x_\nu) & K(x_\mu,\xi_1)\newline K(\xi_1,x_\nu) & K(\xi_1,\xi_1) \end{vmatrix} d\xi_1 \newline +\frac{\lambda^3}{2!}\int_a^b\int_a^b\begin{vmatrix} K(x_\mu,x_\nu) & K(x_\mu,\xi_1) & K(x_\mu,\xi_2)\newline K(\xi_1,x_\nu) & K(\xi_1,\xi_1) & K(\xi_1,\xi_2)\newline K(\xi_2,x_\nu) & K(\xi_2,\xi_1) & K(\xi_2,\xi_2) \end{vmatrix}d\xi_1d\xi_2 - \cdots, $$ and $D_n(x_\mu,x_\mu)\to D(\lambda)$. We arrive at the solution $$ \phi(x) = f(x)+\frac{1}{D(\lambda)}\int_a^bD(x,\xi;\lambda)f(\xi)d\xi. $$ [The kernel $\frac{D(\cdot,\cdot;\lambda)}{D(\lambda)}$ is the resolvent.]