Turn a room's thermostat up on a cold morning. The temperature doesn't jump straight to the target — it climbs, maybe overshoots a little, wobbles, and only after a while settles at a steady reading that barely moves again. Two very different things are happening at once: a short-lived wobble that fades away, and a lasting value the room is heading towards all along. Splitting a system's behaviour into exactly these two pieces — a permanent steady state and a fading transient — is one of the most useful tricks in all of applied mathematics, and it falls straight out of the heat equation.

The heated rod had its ends pinned to zero, so it cooled all the way to 0^\circ. What if the ends are instead held at different fixed temperatures, u(0, t) = T_0 and u(L, t) = T_1? Now the boundary conditions are inhomogeneous, and separation does not apply directly — the sine modes are not zero at both ends.

The fix is a clean split: write the solution as a time-independent steady state plus a decaying transient.

The two pieces

Steady state. As t \to \infty nothing changes, so u_t = 0 and the heat equation reduces to u_s'' = 0. A straight line satisfies that, and the one matching the end temperatures is

Transient. The leftover v = u - u_s satisfies the heat equation with homogeneous ends (v(0,t) = v(L,t) = 0, since u and u_s agree there). That is exactly the problem we already solved: v is a decaying Fourier sine series and dies away. So the rod forgets its starting shape and relaxes onto the steady straight-line profile.

Notice what each equation is missing. The transient's equation still has the time derivative v_t — it is a genuine, evolving heat-flow problem, just with the boring boundary values subtracted off. The steady state's equation u_s'' = 0 has no time in it at all: it is a purely spatial ordinary differential equation. In more than one dimension the very same idea generalises — a steady state that no longer changes with time always solves Laplace's equation, the natural governing law for "settled down and staying put," whether that's temperature in a metal plate or the shape of a stretched membrane.

Worked example: a rod held at 20° and 80°

Take a rod of length L = \pi with its left end clamped at T_0 = 20^\circ and its right end clamped at T_1 = 80^\circ. Finding the steady state takes three short steps, and none of them need Fourier series.

Step 1 — write down the equation. At steady state u_s'' = 0, whose general solution is a straight line, u_s(x) = A + Bx.

Step 2 — fit the boundary values. u_s(0) = A = 20, and u_s(\pi) = 20 + B\pi = 80, so B = 60/\pi.

Step 3 — read off the answer.

That's it — no infinite sum required, because the steady state carries none of the wiggly, mode-by-mode structure that made the original problem hard. Now suppose the rod actually starts out at a uniform 50^\circ everywhere, colder than the right clamp and hotter than the left one. The transient is whatever is left over, v(x, 0) = 50 - u_s(x), and it behaves exactly like the zero-boundary problem from before: a Fourier sine series in x whose coefficients decay like e^{-n^2 t}. However lumpy or lopsided the starting temperature is, that sine series melts away and the rod's profile bends smoothly onto the straight line u_s(x) = 20 + 60x/\pi — never overshooting, never oscillating, just relaxing towards it once and for all.

It's worth spelling out exactly what "the transient is a decaying sine series" means in numbers. Every mode in that series carries its own exponential clock, e^{-n^2 t}, so the fastest-wiggling modes — large n, lots of bumps packed into the rod — die out almost immediately, while the slowest, smoothest mode (n = 1) lingers longest. That's why a rough, lumpy starting temperature first rounds itself off into a gentle single hump before finally straightening into the line u_s: the high-frequency detail vanishes first, the low-frequency trend vanishes last. You can watch exactly that ordering happen — fine structure first, broad shape last — in the chart just below.

Relax to the line

The rod starts cold (u(x, 0) = 0) with its ends suddenly fixed at T_0 and T_1. The faint line is the steady state u_s; the bold curve is u(x, t). Set the end temperatures, then advance time: the transient decays and the profile settles onto the straight line, fastest where the modes are highest.

The thermostat, again

Go back to the heating system that opened this page. Once it switches on, the room's temperature curve looks a lot like the chart above: it climbs, sometimes overshoots the target a little, wobbles briefly — and then flattens out at a constant reading that the thermostat holds from then on. The flat reading it settles at is the steady state: whatever the room's temperature would be if you waited forever with the heater running at a fixed setting. The climbing-and-wobbling part on the way there is the transient: it depends on how cold the room started and exactly how the heater ramps up, and — like the sine series above — it dies away and leaves no trace. Two rooms started at very different temperatures, heated by the same thermostat, end up at the very same steady state; only their transients differ.