Sturm-Liouville Theory

Sturm–Liouville theory is the mathematical spine underneath separation of variables for essentially every linear PDE with variable coefficients on a bounded domain: heat conduction in a non-uniform rod, vibrations of a hanging chain of varying density, and the radial part of the Laplacian in spherical or cylindrical coordinates, where it produces Legendre polynomials and Bessel functions as eigenfunctions. In quantum mechanics, the time-independent Schrödinger equation for a particle in a potential well is literally a Sturm–Liouville eigenvalue problem, and the reality of eigenvalues proved here is the mathematical reason energy levels are real numbers rather than complex ones. Numerical analysts exploit self-adjointness to build stable finite-difference and finite-element discretizations (the resulting stiffness matrices are symmetric, mirroring the continuous operator). In applied statistics and signal processing, weighted orthogonal polynomial families arising from Sturm–Liouville problems (Legendre, Chebyshev, Hermite, Laguerre) underlie Gaussian quadrature rules and orthogonal polynomial regression. The theory also foreshadows Green's Functions for Boundary Value Problems, where the same self-adjoint operator, now inverted rather than diagonalized, produces a symmetric kernel $G(x,\xi)=G(\xi,x)$ directly from the identity proved in part (B) above, and it sets up Eigenfunction Expansions, the general machine for writing arbitrary functions as convergent series in these orthogonal eigenfunctions.

Ask the tutor: How do the boundary-term cancellations in Lagrange's identity change if the Sturm–Liouville problem is posed on a half-line with a decay condition at infinity instead of a second endpoint condition?

Think of the vector space of smooth functions on $[a,b]$ satisfying the boundary conditions as an infinite-dimensional analogue of $\mathbb{R}^n$. The weighted inner product

plays the role of the dot product. The differential expression $L$ plays the role of a matrix $A$. In linear algebra, symmetry of $A$ (meaning $\langle Ax,y\rangle = \langle x, Ay\rangle$ for all $x,y$) is exactly the property that forces real eigenvalues and an orthogonal eigenbasis via the spectral theorem. Sturm–Liouville theory shows that the same symmetry statement, $\langle Lu, v \rangle = \langle u, Lv\rangle$, holds for $L$ acting on functions satisfying the boundary conditions — provided the boundary terms that appear from integrating by parts twice cancel out. That cancellation is precisely what "separated boundary conditions" are built to guarantee. The picture below shows the analogy directly: a symmetric matrix diagonalizes into orthogonal eigenvectors on the left, while the Sturm–Liouville operator diagonalizes into an infinite orthogonal family of eigenfunctions (here, the familiar $\sin(n\pi x)$ family from the Dirichlet problem) on the right.

Example 1 — Robin (radiation) boundary conditions, asymmetric interval. Consider $y'' + \lambda y = 0$ on $[0, L]$ with

Here $p(x)=1$, $q(x)=0$, $w(x)=1$, so this is a regular Sturm–Liouville problem ($p\equiv 1 > 0$, $w\equiv 1>0$ continuous, boundary conditions separated with $(\alpha_1,\alpha_2)=(1,0)$ at $x=0$ and $(\beta_1,\beta_2)=(h,1)$ at $x=L$). By part (C) all eigenvalues are real, so we search over real $\lambda$ and rule out $\lambda \le 0$ first. Case $\lambda \le 0$: write $\lambda = -\mu^2$, $\mu\ge 0$. General solution $y = A\cosh(\mu x) + B\sinh(\mu x)$. The condition $y(0)=0$ forces $A=0$, so $y = B\sinh(\mu x)$. Then $y'(L) + hy(L) = B\big[\mu\cosh(\mu L) + h\sinh(\mu L)\big] = 0$. Since $\mu, h, \cosh, \sinh \ge 0$ with $\cosh > 0$ always, the bracket is strictly positive unless $\mu = 0$ and then it equals $0$ trivially, but $\mu=0$ makes $y\equiv 0$. So no nontrivial solution for $\lambda \le 0$: all eigenvalues are strictly positive, consistent with $q \ge 0$ intuition. Case $\lambda = \mu^2 > 0$: general solution $y = A\cos(\mu x) + B \sin(\mu x)$. The condition $y(0) = 0$ gives $A = 0$, so $y = B\sin(\mu x)$, $B\neq 0$. The right boundary condition:

This transcendental equation has infinitely many positive roots $\mu_1 < \mu_2 < \cdots$ (one in each interval where $\tan(\mu L)$ decreases through $-\mu/h$), giving eigenvalues $\lambda_n = \mu_n^2$ and eigenfunctions $y_n(x) = \sin(\mu_n x)$. By part (D), for $m \neq n$,

even though $\mu_n$ are not the simple multiples $n\pi/L$ of the pure Dirichlet problem — orthogonality here is a structural consequence of self-adjointness, not a lucky trigonometric identity. This is exactly the situation for a rod radiating heat into its surroundings (Newton cooling) at $x=L$.

Example 2 — Legendre's equation on $[-1,1]$: a genuinely singular Sturm–Liouville problem. Legendre's equation,

is the Sturm–Liouville equation $L[y] = \lambda w y$ with $p(x) = 1-x^2$, $q(x) = 0$, $w(x) = 1$, $\lambda = n(n+1)$. Notice $p(\pm 1) = 0$: the regularity hypothesis $p > 0$ on the closed interval fails at both endpoints, so this is a singular Sturm–Liouville problem, not covered directly by the theorem above. Boundedness at $x=\pm1$ replaces a boundary condition. Nevertheless the Lagrange identity computation in part (A) is purely algebraic and does not use positivity of $p$, only that $p \in C^1$. So for any two solutions $u,v$ bounded on $[-1,1]$,

because the factor $(1-x^2)$ itself vanishes at $x=\pm1$, killing the boundary term automatically — no separated boundary condition is even needed, provided $u,v$ and their derivatives stay bounded there (true for the Legendre polynomials $P_n(x)$, which are polynomials, hence bounded, and it is a fact, provable by Frobenius analysis, that any second linearly independent solution blows up logarithmically at $x=\pm1$ and must be excluded). Repeating the argument of part (D) with $u = P_m$, $v = P_n$, $\lambda_m = m(m+1)$, $\lambda_n = n(n+1)$ for $m\neq n$ gives, exactly as before,

Since $m \ne n \Rightarrow m(m+1)\neq n(n+1)$ for nonnegative integers $m,n$ (the map $k\mapsto k(k+1)$ is strictly increasing on $k\ge 0$), we conclude

the classical orthogonality of Legendre polynomials — obtained here without a single integration by parts beyond Lagrange's identity, and without ever invoking the closed-form Rodrigues formula for $P_n$.