Differential Equations

Fourier Series and the Transform

The eigenfunction expansion Sturm-Liouville promised. What becomes of it when the interval opens up to infinity.

Updated July 2026

The Sturm-Liouville notes ended with a promise: a well-posed boundary value problem hands you an orthogonal basis of eigenfunctions, and any reasonable function can be written in it. The oldest and most useful instance of that promise is the one where the operator is just the second derivative and the interval is a finite line. Its eigenfunctions are sines and cosines, and the expansion is the Fourier series.

The simplest eigenproblem

On $[- L, L]$ with periodic boundary conditions, the operator $- d^{2} / d x^{2}$ has eigenfunctions

e^{inπ x / L}, n \in Z,

and they are orthogonal, because $- d^{2} / d x^{2}$ is a Sturm-Liouville operator and its eigenfunctions for distinct eigenvalues must be. Any periodic $f$ then expands as

f (x) = n = - \infty \sum \infty c_{n} e^{inπ x / L}, c_{n} = \frac{1}{2 L} \int_{- L}^{L} f (x) e^{- inπ x / L} d x .

The coefficient formula is just the projection onto one basis vector, the same inner product that gives a component of an ordinary vector. Nothing here is special to sines; it is orthogonality doing the work.

Energy is preserved

Orthonormality gives Parseval's identity, that the total size of a function equals the total size of its coefficients:

\frac{1}{2 L} \int_{- L}^{L} ∣ f (x) ∣^{2} d x = n \sum ∣ c_{n} ∣^{2} .

Physically this is a conservation statement: the energy in a wave is the same whether you count it in space or across its frequency components. Nothing is lost in changing basis.

Letting the interval open up

The Fourier series is tied to a finite interval, which quantises the allowed frequencies into the discrete set $nπ / L$ . Let $L \to \infty$ and those frequencies crowd together into a continuum. The sum becomes an integral and the series becomes the Fourier transform:

\tilde{f} (k) = \int_{- \infty}^{\infty} f (x) e^{- ik x} d x, f (x) = \frac{1}{2 π} \int_{- \infty}^{\infty} \tilde{f} (k) e^{ik x} d k .

Theorem

Why the transform tames differential equations

Under the transform, $d / d x$ becomes multiplication by $ik$ . A linear differential equation with constant coefficients therefore turns into an algebraic one for $\tilde{f} (k)$ , which you solve by division and invert. Differentiation, the hard operation, has been traded for multiplication, the easy one.

The heat equation in one line

Take $\partial_{t} u = D \partial_{x}^{2} u$ and transform in $x$ . The spatial derivative becomes $- k^{2}$ , leaving $\partial_{t} \tilde{u} = - D k^{2} \tilde{u}$ , which is just exponential decay:

\tilde{u} (k, t) = \tilde{u} (k, 0) e^{- D k^{2} t} .

Every mode decays, and the sharp ones, large $k$ , decay fastest. Inverting the transform turns that back into a spreading Gaussian, which is why heat smooths a profile from its roughest features first. The transform did not just solve the equation; it said plainly which part of the initial data survives longest.