Periodic diffusion on a circle ($u(x,t)=\bar{f}+{\sum }_{n\ge 1}(a_n\cos nx+b_n\sin nx)e^{-D{n}^{2}t}$)

For a circle of circumference $2\pi$, the heat equation $u_t=D{u}_{xx}$ has periodic matching of both $u$ and $u_x$ at $x=\pm \pi$. For $D>0$, the displayed Fourier series solves the initial-value problem, with $\bar{f}=(2\pi {)}^{-1}{\int }_{-\pi }^{\pi }f(x)dx$ and the usual real Fourier coefficients. Every nonconstant normal mode decays, leaving the conserved mean. In particular $f(x)=(\pi -|x|{)}^2$ gives $\bar{f}={\pi }^2/3$, $a_n=4/n^2$ and $b_n=0$. The initial cusp is smoothed for every positive time.

For $L>0$, extend $f_L(x)=\sin x-x\sin L/L$ from $[-L,L]$ periodically. The linear term makes both endpoint values zero and the endpoint first derivatives match. The function is odd, so its Fourier series has only sine terms, with the displayed Fourier coefficients. Indeed, for $k_n=n\pi /L\ne 1$,subtracting and multiplying by $2/L$ proves the formula. The coefficients decay as $O(n^{-3})$. At $L=m\pi$, the single coefficient $b_m$ has removable limit $1$, with all others zero. Multiplying $b_n$ by $e^{-\kappa k_n^{2}t}$ solves the corresponding heat equation on the circle.