Laplace on a Rectangle
The classic boundary-value problem: find the steady temperature inside a square plate when its
edges are held fixed. On [0, \pi] \times [0, \pi], suppose three sides
are kept at 0 and the top edge at u(x, \pi) = f(x).
Solve u_{xx} + u_{yy} = 0 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.
- Separation gives \sin(nx) across the zero side-edges and \sinh(ny) rising to the hot edge.
- The solution is \sum b_n \frac{\sinh(ny)}{\sinh(n\pi)}\sin(nx), with b_n the Fourier sine coefficients of the edge data.
- Hyperbolic functions appear in the unbounded direction; trigonometric ones across the zero-edges.
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.