The heat equation on an interval

Separation of variables and Fourier series combine into a complete solution of \( u_t = k\,u_{xx} \), \( u(0,t) = u(L,t) = 0 \), \( u(x,0) = f(x) \): \[ u(x,t) = \sum_{n=1}^\infty b_n\,e^{-k n^2\pi^2 t/L^2}\sin\frac{n\pi x}{L}, \qquad b_n = \frac2L\int_0^L f(x)\sin\frac{n\pi x}{L}\,dx. \] The sine coefficients on \( [0, L] \) are the Fourier coefficients of the odd extension of \( f \). For \( f = 1 \) on \( [0, 1] \), \( b_n = 2(1 - (-1)^n)/(n\pi) \): the example computes \( b_1 = 4/\pi \), and every even coefficient vanishes.

Long time. Every term decays, the first most slowly, so \( u(x,t) \approx b_1 e^{-k\pi^2 t/L^2}\sin(\pi x/L) \) once \( t \) is a few multiples of \( L^2/(k\pi^2) \). The decay rate is the smallest eigenvalue, which doubling the length of the rod divides by four.

Instant smoothing. Even if \( f \) jumps, \( |b_n| \) is bounded, and for any fixed \( t > 0 \) the factors \( e^{-kn^2\pi^2t/L^2} \) decay faster than any power of \( n \). Differentiating the series term by term any number of times still gives uniformly convergent series, so \( u(\cdot, t) \) is infinitely differentiable for every \( t > 0 \). With insulated ends, \( u_x(0,t) = u_x(L,t) = 0 \), use cosines instead: the constant mode has eigenvalue zero, the mean temperature \( \frac1L\int_0^L f \) is conserved, and the rod relaxes to it. Fixed non-zero end temperatures are handled by subtracting the linear steady state first.

Picture it: a rod heated uniformly and plunged into ice at both ends. Its flat temperature profile rounds off at the corners immediately, becomes a sine arch within moments, and the arch then sinks towards zero while keeping its shape.

Think it: run the formula backwards and the factors become \( e^{+kn^2\pi^2 t/L^2} \). A tiny high-frequency error in the final state is amplified beyond any bound, so the backward heat equation has no stable solution for general data. Diffusion destroys information, and a well-posed problem can only run in the direction that destroys it.