Regular Sturm-Liouville Problems

Definition A second-order eigenvalue equation is in self-adjoint form, also called Sturm-Liouville form, when it reads

where the functions $p$, $q$ and $w$ of $x$ are the leading coefficient, the potential and the weight. Here $p$ sits inside the outer derivative, so it is not the middle coefficient of the normalized form $y'' + p\,y' + q\,y$ used for initial-value problems. Write $L[y] = -(p\,y')' + q\,y$. For a continuously differentiable $p$ and twice continuously differentiable $u$ and $v$, the Lagrange identity

holds, because the integrand is exactly the derivative of $p\,(uv' - u'v)$. The operator $L$ is self-adjoint on a set of functions when this boundary term vanishes for every pair $u$, $v$ in the set. Any equation $a_2y'' + a_1y' + a_0y + \lambda\,a_3\,y = 0$ with continuous coefficients and $a_2 > 0$ on $[a, b]$ can be brought into this form. Multiply it by $p/a_2$, where

which gives $q = -p\,a_0/a_2$ and $w = p\,a_3/a_2$.

Definition The coefficient hypotheses of a regular problem on a finite closed interval $[a, b]$ are that $p$ is continuously differentiable on $[a, b]$, that $q$ and $w$ are continuous on $[a, b]$, that all three are real, and that

Positivity of $p$ at every point, the endpoints included, means the equation can be divided by $p$, and the normalized equation

has continuous coefficients on all of $[a, b]$. So each initial-value problem at a point of $[a, b]$ has exactly one solution there. Positivity of $w$ makes $\int_a^b w\,u^2\,dx > 0$ for every continuous $u$ that is not identically zero, so the weighted integral below is a genuine inner product. These hypotheses say nothing about the sign of $q$, and they do not make the eigenvalues positive.

Definition Separated boundary conditions impose one condition at each end,

with real constants such that $(\alpha_1, \beta_1) \ne (0, 0)$ and $(\alpha_2, \beta_2) \ne (0, 0)$. The first condition involves only the left endpoint and the second only the right endpoint. Dirichlet, Neumann and Robin conditions are all of this kind. If $u$ and $v$ both satisfy the left condition, then the vectors $(u(a), u'(a))$ and $(v(a), v'(a))$ both solve the single equation $\alpha_1 s + \beta_1 t = 0$ with $(\alpha_1, \beta_1) \ne (0, 0)$, so they are parallel and

The same holds at $b$, so the boundary term in the Lagrange identity vanishes, and $L$ is self-adjoint on the functions that satisfy the conditions.

Definition A regular Sturm-Liouville problem consists of the equation

on a finite closed interval, with the coefficient hypotheses above, which require $p > 0$ and $w > 0$ on all of $[a, b]$, and with real separated boundary conditions at both ends. A number $\lambda$, real or complex, is an eigenvalue when the problem has a solution that is not identically zero, and that solution is an eigenfunction. Problems on an infinite interval, or with $p$ or $w$ vanishing at an endpoint, are called singular and need separate treatment.

Definition For real functions on $[a, b]$ the weighted inner product is

In a regular Sturm-Liouville problem, real eigenfunctions $u$ and $v$ with different eigenvalues $\lambda \ne \mu$ satisfy $\langle u, v \rangle_w = 0$. To prove it, put $L[u] = \lambda w u$ and $L[v] = \mu w v$ into the Lagrange identity. The $q$ terms cancel, and the separated conditions make the boundary term zero, so

and the integral vanishes because $\lambda \ne \mu$. If $f = \sum_n c_n y_n$ is an expansion in eigenfunctions with distinct eigenvalues, taking its weighted inner product with a fixed $y_m$ removes every term except the one with $n = m$, which gives the coefficients

and an eigenfunction is normalized when $\langle y_m, y_m \rangle_w = 1$. The eigenfunctions of a regular problem are also complete. For every piecewise continuous $f$ on $[a, b]$, the series with these coefficients converges to $f$ in the weighted mean-square sense, which by itself does not give convergence at every point.

Definition For a regular Sturm-Liouville problem the following statements hold.

1. Every eigenvalue is real, and its eigenfunctions can be taken real-valued.
2. Every eigenvalue is simple, which means that all of its eigenfunctions are multiples of one of them, so each eigenspace has dimension $1$.
3. The eigenvalues form an infinite increasing sequence $\lambda_1 < \lambda_2 < \lambda_3 < \cdots$ with $\lambda_n \to \infty$. There is a smallest eigenvalue, but it need not be positive.

The first two statements follow from the structure above. If $L[y] = \lambda w y$ with $y$ not identically zero, then the complex conjugate $\bar{y}$ satisfies the same real boundary conditions and $L[\bar{y}] = \bar{\lambda} w \bar{y}$, because $p$, $q$ and $w$ are real. The Lagrange identity with $u = y$ and $v = \bar{y}$ then gives

and the integral is positive, so $\lambda$ is real. The real and imaginary parts of $y$ then solve the same problem, and at least one of them is not identically zero. For simplicity, the condition at $a$ forces the pair $(y(a), y'(a))$ of every eigenfunction to be a multiple of $(\beta_1, -\alpha_1)$. Two eigenfunctions for the same $\lambda$ therefore have proportional initial data at $a$, and uniqueness for the initial-value problem makes them proportional on all of $[a, b]$. The third statement is a deeper part of the same theorem.