Reduction of a boundary value problem to a Fredholm equation

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.

Going deeper

For a student

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.