Sturm–Liouville problem as an integral equation

Example

Consider the simplest Sturm–Liouville problem: find $\lambda$ and a nonzero $u \in C^2[0, 1]$ with $$-u''(x) = \lambda u(x), \quad u(0) = u(1) = 0.$$ The Green's function of $-u''$ with these boundary conditions is $$G(x, s) = \begin{cases} s(1 - x), & 0 \le s \le x \le 1, \\ x(1 - s), & 0 \le x \le s \le 1. \end{cases}$$ For continuous $f$, the unique solution of $-u'' = f$, $u(0) = u(1) = 0$ is $u(x) = \int_0^1 G(x, s) f(s) \, ds$. Taking $f = \lambda u$, the boundary value problem becomes the homogeneous Fredholm equation of the second kind $$u(x) = \lambda \int_0^1 G(x, s) \, u(s) \, ds.$$ Conversely, a continuous solution of this integral equation is $C^2$ and solves the boundary value problem.

Check: the kernel $G$ is continuous and symmetric. For $u_n(x) = \sin(n \pi x)$, the function $v = u_n / (n^2 \pi^2)$ satisfies $-v'' = u_n$ and $v(0) = v(1) = 0$. So $\int_0^1 G(x, s) \sin(n \pi s) \, ds = \sin(n \pi x) / (n^2 \pi^2)$, and the characteristic values of $G$ are $\lambda_n = n^2 \pi^2$, $n \ge 1$. These are exactly the eigenvalues of the boundary value problem.