---
title: "Sturm-Liouville Problems and Eigenfunction Expansions"
description: "Putting a second-order boundary value problem in self-adjoint form makes its eigenfunctions orthogonal, which is why a solution can be written as a series in them at all. Fourier series is then one ca"
canonical: https://lightmysky.com/learn/mathematics/sturm-liouville-problems-and-eigenfunction-expansions-mt_oqadAPaSsW
source: https://lightmysky.com/learn/mathematics/sturm-liouville-problems-and-eigenfunction-expansions-mt_oqadAPaSsW.md
retrieved: 2026-09-12
---

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# Sturm-Liouville Problems and Eigenfunction Expansions

Putting a second-order boundary value problem in self-adjoint form makes its eigenfunctions orthogonal, which is why a solution can be written as a series in them at all. Fourier series is then one case rather than a trick that happened to work.

Subject: Mathematics · Area: Differential Equations · Ages 21 to 22
Page: https://lightmysky.com/learn/mathematics/sturm-liouville-problems-and-eigenfunction-expansions-mt_oqadAPaSsW

## Ready when they can

- Writes a given second-order equation in Sturm-Liouville form and identifies the weight function
- Shows that eigenfunctions for distinct eigenvalues are orthogonal with respect to that weight
- Expands a given function in the eigenfunctions of a stated boundary value problem and gets the coefficients

## Lesson: One framework for every vibrating string

A boundary value problem adds conditions at both ends of an interval instead of at one starting point. Only special values of the parameter admit nonzero solutions that satisfy both ends. Those values are the eigenvalues and the matching solutions are the eigenfunctions. Sturm-Liouville theory is the clean framework that organizes all of them at once. For y double prime plus w y equals 0 with y(0) = 0 and y(pi) = 0, the eigenvalues are 1, 4, 9, and so on.

**Example.** Take the equation y'' + (2/x) y' + w y = 0, where w plays the role of the eigenvalue. Multiply through by x squared to get x^2 y'' + 2x y' + w x^2 y = 0. The first two terms collapse into exactly (x^2 y')' by the product rule. The standard form reads (p y')' + (w times weight minus q) y = 0, so the weight function here is x squared. You already know the integrating factor trick from first-order equations: this is the same move in reverse.

Eigenfunctions for distinct eigenvalues are orthogonal with respect to that weight: the integral of their product times the weight is zero. Different modes do not overlap. Sines on an interval behave this way, and so do the Legendre pair 1 and x on minus one to one, whose product integrates to zero. Boundary conditions at each end are what pick out the allowed eigenvalues.

**Tip.** To expand a function, copy the power series idea with modes as building blocks. Multiply by one eigenfunction and the weight, integrate, and divide by its squared norm: every other mode vanishes, leaving a single coefficient. When the modes are sines and cosines you get Fourier series as the famous special case. Taylor series match derivatives at one point, while these expansions match across the whole interval.

**Recap.** Self-adjoint form reveals the weight, the weight makes modes orthogonal, and orthogonality hands you each coefficient alone.

## Practice

14 questions on this page, each with its working shown.

## Needs first

- [Fourier Series](https://lightmysky.com/learn/mathematics/fourier-series-mt_DJ3iwfj7NK)
- [Second-Order Linear Equations with Constant Coefficients](https://lightmysky.com/learn/mathematics/second-order-linear-equations-with-constant-coefficients-mt_G7sMq1EHwQ)

## Opens up

- [Green's Functions for Boundary Value Problems](https://lightmysky.com/learn/mathematics/greens-functions-for-boundary-value-problems-mt_2ctFrfKIgW)
