---
tag: 6Q36P44PHZ
kind: statement
role: proposition
name: Reduction of a boundary value problem to a Fredholm equation
msc: [45B05, 34B05, 34B27]
uses: [TM8VQJWGHZ, SH6JY9RN2E, 526TCVMSHP, 8KJCN49V64]
about: [526TCVMSHP, TM8VQJWGHZ]
sources:
  - url: https://en.wikipedia.org/wiki/Green%27s_function
---

## Statement

Let $a < b$ and let $q, F \colon [a, b] \to \R$ be continuous.
Let $G \colon [a, b] \times [a, b] \to \R$ be the [[Green's function|SH6JY9RN2E]] of $y'' = 0$ with $y(a) = y(b) = 0$, namely
$$G(x, s) = \begin{cases} \dfrac{(x - a)(s - b)}{b - a}, & a \le x \le s \le b, \\[2mm] \dfrac{(s - a)(x - b)}{b - a}, & a \le s \le x \le b. \end{cases}$$
For a continuous function $y \colon [a, b] \to \R$, the following are equivalent.

1. $y \in C^2[a, b]$ and $y$ solves the [[boundary value problem|TM8VQJWGHZ]] $y'' + q(x) y = F(x)$, $y(a) = y(b) = 0$.
2. $y$ solves the [[Fredholm integral equation|526TCVMSHP]] of the second kind
$$y(x) = \int_a^b G(x, s) F(s) \, ds - \int_a^b G(x, s) q(s) y(s) \, ds, \quad x \in [a, b].$$

## Remarks

Proof idea: for continuous $g$, the function $w(x) = \int_a^b G(x, s) g(s) \, ds$ is the unique $C^2$ solution of $w'' = g$, $w(a) = w(b) = 0$; uniqueness holds because a linear function vanishing at $a$ and $b$ is zero.
If $y$ satisfies 1, apply this with $g = F - q y$ to get 2.
If $y$ satisfies 2, the right side is $w$ with the continuous function $g = F - q y$, so $y = w$ is $C^2$ with $y'' = F - q y$ and $y(a) = y(b) = 0$.

Example: on $[0, 1]$, $G(x, s) = x (s - 1)$ for $x \le s$ and $s (x - 1)$ for $x \ge s$.
The eigenvalue problem $y'' + \mu y = 0$, $y(0) = y(1) = 0$, becomes the homogeneous equation $y(x) = -\mu \int_0^1 G(x, s) y(s) \, ds$, whose kernel $-G$ is symmetric.
Its nonzero solutions occur for $\mu = k^2 \pi^2$, $k \ge 1$, with $y = \sin(k \pi x)$.

The integral equation has the continuous kernel $-G(x, s) q(s)$, so Fredholm theory applies.
So the boundary value problem has a unique solution for every $F$ if and only if the problem with $F = 0$ has only the zero solution.
The same method works for other separated boundary conditions and other second-order operators, provided the Green's function of the unperturbed problem exists.

## Explanations

### L1 human

A boundary value problem fixes the values of the solution at both ends of an interval.
The Green's function turns the differential equation into an integral equation in which the boundary values are already built in.
The resulting equation integrates over the whole interval, so it is a Fredholm equation rather than a Volterra equation.

### L2 human

This reduction is how Sturm–Liouville problems enter the theory of integral equations.
For self-adjoint problems the Green's function is a symmetric kernel, and the Hilbert–Schmidt theory of symmetric kernels then gives real eigenvalues and expansions in eigenfunctions.
In contrast, an initial value problem leads to a Volterra equation, which is always uniquely solvable.
A boundary value problem leads to a Fredholm equation, which may fail to be solvable at characteristic values.