Nonzero Boundary Temperatures

A nonzero boundary condition simply means the ends of the rod are held at some temperature other than 0°, say one end in boiling water and the other in a 20° room. That's the usual situation in real life, so we need a way to handle it. The trick is to write: temperature now = where the rod will end up + how far it still has to go. In $u = u_E + v$ there are just two pieces: $u_E$ (the steady temperature the rod settles into, a straight line between the end temperatures) and $v$ (the leftover gap, which shrinks to nothing over time). Two simple pieces, that's all.

Why the subtraction works. At $x = 0$, $u = T_1$ and $u_E = T_1$, so $v = 0$, and the same thing happens at $x = L$. And since $u_E'' = 0$ and $u_E$ doesn't depend on $t$, it satisfies the heat equation on its own, so subtracting it leaves the equation unchanged. So $v$ is a zero-boundary problem, and everything from the last two lessons works on it. If that feels familiar, it should: it's the same idea as $y = y_c + y_p$ for ODEs, with $u_E$ playing the part of the particular solution that handles the "forcing" from the ends.

The steps:

1. Find the equilibrium: $u_E'' = 0$ with $u_E(0) = T_1$, $u_E(L) = T_2$.
2. Let $v = u - u_E$. Then $v_t = k\,v_{xx}$ with $v(0, t) = v(L, t) = 0$ and $v(x, 0) = f(x) - u_E(x)$.
3. Solve for $v$ by separation of variables and add back $u_E$.

As $t \to \infty$, $v \to 0$ and $u \to u_E$.

A common mistake. It's tempting to use the original starting temperature $f(x)$ as the initial condition for $v$. It isn't: $v$ measures the gap from equilibrium, so it starts at $f(x) - u_E(x)$. A rod that starts at 0° everywhere, for instance, gives $v(x, 0) = -u_E(x)$, not $0$.

Heat sources. With a steady source $Q(x)$, the equilibrium solves $k\,u_E'' = -Q$ instead, giving a curved steady profile, and the same subtraction works.

Heat exchangers, building walls between indoor and outdoor temperatures, and the temperature inside the Earth's crust are all equilibrium-plus-transient problems.