---
tag: 4BY024AVR8
kind: statement
role: proposition
name: Solution of equations with degenerate kernels
msc: [45B05, 45A05]
uses: [6Y1RY5767R, DEYYNVCV9J, 00KDZRW4C0]
about: [6Y1RY5767R, DEYYNVCV9J]
sources:
  - url: https://en.wikipedia.org/wiki/Fredholm_integral_equation
---

## Statement

Let $a < b$, let $m \ge 1$, and let $a_1, \dots, a_m, b_1, \dots, b_m \in L^2[a, b]$.
Let $K(x, y) = \sum_{i=1}^{m} a_i(x) b_i(y)$ be the corresponding [[degenerate kernel|6Y1RY5767R]], let $\lambda \in \C$ and $f \in L^2[a, b]$.
Let $A$ be the $m \times m$ matrix and $F \in \C^m$ the vector with entries
$$A_{ij} = \int_a^b b_i(y) \, a_j(y) \, dy, \qquad F_i = \int_a^b b_i(y) \, f(y) \, dy.$$
Consider the [[Fredholm equation of the second kind|DEYYNVCV9J]]
$$\varphi(x) - \lambda \int_a^b K(x, y) \varphi(y) \, dy = f(x).$$
Its solutions $\varphi \in L^2[a, b]$ are exactly the functions $\varphi = f + \lambda \sum_{i=1}^{m} c_i a_i$, where $c \in \C^m$ solves $(I_m - \lambda A) c = F$.
Distinct solutions $c$ give distinct functions $\varphi$.
In particular, if $\det(I_m - \lambda A) \ne 0$, the integral equation has exactly one solution for every $f$.

## 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|00KDZRW4C0]] that governs the problem; it equals the Fredholm determinant of $K$ when $K$ is continuous.

## Explanations

### L1 human

When the kernel is a finite sum of products of a function of $x$ and a function of $y$, the integral equation hides only finitely many unknowns.
These unknowns are the numbers $\int b_i \varphi$, and they satisfy an ordinary square linear system.
Solve the system, and the formula gives the solution of the integral equation.

### L2 human

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.