Steady State and Transient

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.

u(x, t) = u_s(x) + v(x, t).

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

u_s(x) = T_0 + (T_1 - T_0)\frac{x}{L}.

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.

Split u = u_s(x) + v(x, t) to absorb inhomogeneous boundary data into u_s.
u_s solves u_s'' = 0 with the given end values — a straight line for the 1-D rod.
v solves the heat equation with homogeneous ends — a decaying sine series.
As t \to \infty, v \to 0 and u \to u_s.

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.