Laplace on a Rectangle

This is the payoff moment. You've built separation of variables and the Fourier series machinery to solve time-dependent problems like the heated rod. Now watch the exact same toolkit crack a problem with no time in it at all: Laplace's equation on a rectangular plate. Find the steady temperature inside a square metal sheet once its four edges are clamped at fixed temperatures and everything inside has settled down for good. It's a genuinely satisfying full-circle moment: the same sines, the same coefficient-matching trick, applied to a completely different kind of problem.

The setup, in its cleanest form: on the square [0, \pi] \times [0, \pi], hold three sides at 0 and let the top edge carry a specified profile u(x, \pi) = f(x). That looks like a very special case — real plates rarely have three edges sitting at exactly zero — but it's a standard, deliberate simplification. Once you can solve this one problem, the superposition principle lets you build the general four-hot-edges plate for free: solve the "only the top edge is nonzero" problem four times, once with each edge playing the "hot" role in turn, and add the four solutions together. Solve u_{xx} + u_{yy} = 0 for this one-edge case by separation of variables.

Separation, with a twist in y

Seek u = X(x)Y(y). Separation gives X'' = -\lambda X and Y'' = +\lambda Y — note the opposite sign. The zero conditions on the left and right edges make X the familiar sines \sin(nx) with \lambda = n^2. But the Y equation now has hyperbolic solutions \sinh(ny), \cosh(ny); the zero condition on the bottom edge picks \sinh(ny).

Superposing and matching the top edge,

u(x, y) = \sum_{n=1}^{\infty} b_n\,\frac{\sinh(ny)}{\sinh(n\pi)}\,\sin(nx),

where the b_n are the Fourier sine coefficients of f. The \sinh(ny)/\sinh(n\pi) factor is 1 on the hot top edge and decays into the cool interior — the boundary heat soaks inward exactly as intuition demands.

Worked example: how fast does the heat fade with distance?

Take the simplest possible edge profile, f(x) = \sin(x), so only the n = 1 term survives and b_1 = 1. The solution collapses to

u(x, y) = \frac{\sinh(y)}{\sinh(\pi)}\sin(x).

Plug in three heights and watch the \sinh(y)/\sinh(\pi) factor — the fraction of the boundary heat that survives at height y:

Each step down roughly multiplies the remaining fraction by the same small factor — exactly what an exponential decay looks like (indeed \sinh(y) \approx e^{y}/2 once y is away from 0, so the ratio \sinh(y)/\sinh(\pi) behaves like e^{y - \pi}). This matches plain physical intuition: a point deep in the interior, far from the one hot edge, should barely feel it — and the maths says precisely how fast that feeling fades.

Heat soaking in from one edge

The plate's steady temperature with the top edge held at u = 1 and the other three at 0. Warm shades crowd against the top and fade smoothly downward — no interior hot spot, just a gentle harmonic gradient set entirely by the boundary.

Match the picture against the numbers from the worked example above: the warm band hugging the top edge is exactly that \approx 100\%-strength region near y = \pi, the mid-height band has already faded to roughly the 20\% level computed for y = \pi/2, and by the time you reach the bottom edge the colour has washed out almost completely, consistent with the boundary condition u(x, 0) = 0 forcing it all the way to zero. The full sum over every n looks a little richer than the single-mode estimate — higher harmonics bend the contours near the corners — but the overall exponential fade with depth is unmistakable at a glance.

It's tempting to reach for the same recipe you used on the heated rod: sines multiplying a decaying exponential in time, \sin(nx)\,e^{-\alpha n^2 t}. That pattern was correct there — but it simply cannot appear here, because Laplace's equation has no t at all. There is nothing to decay over time. Instead, the second spatial direction pairs sines with hyperbolic sine and cosine: \sinh(ny) and \cosh(ny). These look deceptively similar to their circular cousins — same names, same-looking symbols — but they behave completely differently: \sinh and \cosh grow (roughly) exponentially rather than oscillating, and neither one decays anywhere on its own. It's an easy slip to write "sinh" when you mean "sin" (or vice versa) halfway through a calculation — always check which boundary condition forced which choice: the zero conditions gave the trigonometric functions, and the direction with no natural period (the one running off to the hot edge) gave the hyperbolic ones.

This rectangular hot-plate problem isn't a textbook invention dreamed up for practice — it's essentially the exact question that launched the whole subject. In the early 1800s, Jean-Baptiste Joseph Fourier was studying heat flow in solid bodies (partly motivated by, of all things, questions about the Earth's internal temperature) and needed to represent an arbitrary boundary temperature profile as a sum of sines — the object we now call a Fourier series. Pierre-Simon Laplace, his contemporary and sometime rival at the French Academy, was simultaneously working out the equation for gravitational and electrostatic potentials that now carries his name. Bolt the two together — Fourier's series expansions plus Laplace's equation — and you get precisely the rectangle-plate solution above. A thoroughly practical 19th-century engineering question — how does heat spread through a solid object? — ended up seeding some of the deepest and most widely used pure mathematics of the next two centuries, from signal processing to quantum mechanics.

The two men weren't distant strangers, either. Laplace sat on the very examining committee that reviewed Fourier's 1807 memoir on heat, and reportedly wasn't fully convinced at first that an arbitrary function could really be built out of infinitely many sines and cosines — a doubt that took years, and other mathematicians' scrutiny, to fully settle. It's a nice reminder that even the giants whose names now sit side by side on this very page once argued over whether the maths in front of you was even valid.