Insulated Ends and Circular Rings

An insulated end is simply one that heat can't get through, like the end of a rod wrapped in thick foam. Mathematically, no heat flow means a flat temperature there ($u_x = 0$, zero slope), by Fourier's law. A rod sealed at both ends keeps all its heat and just spreads it around until it's even, which is how a thermos or a well-lagged pipe behaves. The formula has three pieces: $A_0$ (the average starting temperature, which never changes), $A_n$ (how much of each cosine wave was there at the start), and the fading factors $e^{-k(n\pi/L)^2t}$ (which shrink those waves away). So in the end only $A_0$ is left. It's not as bad as it looks: it's the last lesson with cosines in place of sines, plus one constant.

Insulated ends. The eigenvalue problem $X'' + \lambda X = 0$, $X'(0) = X'(L) = 0$ has $\lambda_0 = 0$ with $X_0 = 1$ (a constant has zero slope everywhere) and $\lambda_n = \left(\frac{n\pi}{L}\right)^2$ with $X_n = \cos\frac{n\pi x}{L}$. Why cosines? Insulated means zero slope at the ends, and cosines are flat at $0$ and $L$ while sines aren't. That's the table from the Fourier toolkit lesson in action.

Conservation. The constant mode never decays, and it equals the average of $f$: total heat is conserved, so the final uniform temperature is the initial average. This is the big difference from fixed ends. There, the ice baths kept pulling heat out until everything hit 0°. Here the heat has nowhere to go, so it just spreads out.

A common mistake. With fixed ends, $\lambda = 0$ gave nothing, so it's tempting to skip it here too. Don't: with insulated ends, $\lambda = 0$ gives the constant $X_0 = 1$, and dropping it throws away the final temperature entirely.

The ring. A ring has no ends at all, so the only requirement is that the temperature and its slope match up where the two ends of the opened-out wire meet. A thin circular wire, opened out to $-L \le x \le L$, needs $u(-L, t) = u(L, t)$ and $u_x(-L, t) = u_x(L, t)$. Every eigenvalue $\left(\frac{n\pi}{L}\right)^2$ now has two eigenfunctions, a cosine and a sine, and the coefficients are those of the full Fourier series on $[-L, L]$.

Insulated-end models describe a sealed thermos or a well-lagged pipe; the ring problem was Fourier's own example when he invented Fourier series in 1807.