Differential Equations

Separation of Variables

Assuming the solution factorises works far more often than it has any right to. The coordinate system is why.

Updated May 2026

Almost every partial differential equation solved by hand is solved the same way: guess that the solution factorises, and see whether the equation permits it.

The method

Take the heat equation on a bar of length $L$ , held at zero at both ends:

\frac{\partial u}{\partial t} = α \frac{\partial ^{2} u}{\partial x ^{2}}

Try $u (x, t) = X (x) T (t)$ . Substituting and dividing through by $X T$ :

\frac{1}{α T} \frac{d T}{d t} = \frac{1}{X} \frac{d ^{2} X}{d x ^{2}}

The left side depends only on $t$ , the right only on $x$ . Two functions of independent variables can agree everywhere only by both being constant. Call it $- λ$ and the PDE becomes two ODEs.

The boundary conditions do the real work

$X^{''} + λ X = 0$ with $X (0) = X (L) = 0$ has solutions only for

λ_{n} = (\frac{nπ}{L})^{2}, X_{n} = sin \frac{nπ x}{L}

This is an eigenvalue problem, and the discreteness is imposed by the boundary, not by the equation. A bounded domain quantises. That single observation is the entire origin of quantised energy levels in quantum mechanics. The Schrödinger equation is a boundary value problem and the integers come from the box.

Fig. 1

The first three modes of a fixed string. Each has an integer number of half-wavelengths in the interval, which is what the boundary conditions force. The dashed curve is the same mode half a period later.

Superposing and matching the initial condition:

u (x, t) = n = 1 \sum \infty b_{n} sin \frac{nπ x}{L} e^{- α λ_{n} t}

with $b_{n}$ the Fourier sine coefficients of $u (x, 0)$ .

Why high modes vanish first

Each term decays at rate $α λ_{n} \propto n^{2}$ . Fine structure, meaning large $n$ , disappears quadratically faster than coarse structure.

This is why heat diffusion smooths, and why running it backwards is hopeless: the inverse problem amplifies exactly the modes that carry the least reliable information. It is the same reason deblurring an image is hard.

When it fails

Separation requires the boundary to be a coordinate surface. A rectangle in Cartesians, a disc in polars, a sphere in sphericals are all fine. An L-shaped domain, or a non-concentric annulus, is not: no coordinate system separates it, and no amount of cleverness will make it.

There are exactly eleven coordinate systems in which the Helmholtz equation separates. Everything else needs conformal mapping, Green's functions, or a computer.