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Separation of variables gives $X^{\prime \prime }=-k^{2}X$ and $T^{\prime }=-\kappa k^{2}T$. The periodic matching of both the function and its first derivative selects $k=n\pi /L$, with the constant mode at $n=0$. Thus the periodic diffusion on a circle solution iswhere the Fourier coefficients of the initial function areThe formula for $a_n$ also gives $a_0$ at $n=0$. The constant mode preserves the initial spatial mean; every nonconstant Fourier mode decays. For initial data in L2 space, the series approaches those data in L2 space as $t\downarrow 0$ and is smooth at positive times.

Since the endpoints represent the same point of the tube, both density and diffusive flux must match. For a positive diffusion coefficient $\lambda$, the periodic boundary conditions areFor separation of variables, periodic spatial modes solve $X^{\prime \prime }+n^{2}X=0$ and have time factors $e^{-\lambda n^{2}t}$. The constant mode has no decay. The periodic diffusion on a circle solution is thereforewhere the Fourier coefficients areThe expression for $a_n$ includes $n=0$.

The specified profile is even, so $b_n=0$. Its mean follows from ${\int }_0^{\pi }(\pi -x{)}^{2}dx={\pi }^3/3$, giving $a_0=2{\pi }^2/3$. For $n\ge 1$, two integrations by parts giveConsequentlyAll nonconstant normal modes decay for $\lambda >0$, so the density becomes uniform:The conserved total amount is ${\int }_{-\pi }^{\pi }udx=2{\pi }^3/3$. The initial cusp at zero is compatible with an L2 space solution and is immediately smoothed by diffusion. At $\lambda =0$ there would be no relaxation, so positivity is needed for this long-time conclusion.