Solution of equations with degenerate kernels

Remarks

If $\varphi$ is a solution, put $c_i = \int_a^b b_i(y) \varphi(y) \, dy$. The equation says $\varphi = f + \lambda \sum_i c_i a_i$, and multiplying this by $b_i$ and integrating gives $c_i = F_i + \lambda \sum_j A_{ij} c_j$. Conversely, if $c$ solves the system and $\varphi = f + \lambda \sum_i c_i a_i$, the same computation shows $\int_a^b b_i \varphi = c_i$, hence $\lambda \int_a^b K(\cdot, y) \varphi(y) \, dy = \lambda \sum_i c_i a_i = \varphi - f$. This also recovers $c$ from $\varphi$, which gives the injectivity.

Example: on $[0, 1]$ with $K(x, y) = xy$, so $m = 1$ and $a_1(x) = b_1(x) = x$, one gets $A = (1/3)$. For $\lambda = 1$ and $f(x) = x$, $F = 1/3$, so $c = (1/3)/(2/3) = 1/2$ and $\varphi(x) = 3x/2$. For $\lambda = 3$ the system is $0 \cdot c = F$, solvable only when $F = \int_0^1 y f(y) \, dy = 0$.

The polynomial $\det(I_m - \lambda A)$ is the determinant that governs the problem; it equals the Fredholm determinant of $K$ when $K$ is continuous.

Going deeper

For a student

Degenerate kernels are the base case of Fredholm theory. Every square-integrable kernel can be approximated by degenerate kernels, and the general theorems are proved by reducing to this case plus a small remainder that a Neumann series handles. The same reduction is a practical numerical method: expand the kernel in a few separable terms and solve a small linear system.