= Solution
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> are
$$
\boxed{u(-\pi,t)=u(\pi,t),\qquad u_x(-\pi,t)=u_x(\pi,t).}
$$
For <separation of variables>, periodic spatial modes solve $X''+n^2X=0$ and have time factors $e^{-\lambda n^2t}$. The constant mode has no decay. The <periodic diffusion on a circle> solution is therefore
$$
u(x,t)=\frac{a_0}{2}+\sum_{n=1}^\infty[a_n\cos(nx)+b_n\sin(nx)]e^{-\lambda n^2t},
$$
where the <Fourier coefficients> are
$$
a_n=\frac1\pi\int_{-\pi}^\pi f(x)\cos(nx)dx,\qquad b_n=\frac1\pi\int_{-\pi}^\pi f(x)\sin(nx)dx.
$$
The expression for $a_n$ includes $n=0$.
The specified profile is even, so $b_n=0$. Its mean follows from $\int_0^\pi(\pi-x)^2dx=\pi^3/3$, giving $a_0=2\pi^2/3$. For $n\geq1$, two <integrations by parts> give
$$
\int_0^\pi(\pi-x)^2\cos(nx)dx=\frac{2\pi}{n^2},\qquad a_n=\frac4{n^2}.
$$
Consequently
$$
\boxed{u(x,t)=\frac{\pi^2}{3}+4\sum_{n=1}^\infty\frac{\cos(nx)}{n^2}e^{-\lambda n^2t}.}
$$
All nonconstant <normal modes> decay for $\lambda>0$, so the density becomes uniform:
$$
\boxed{u(x,t)\longrightarrow\frac{\pi^2}{3}\quad(t\to\infty).}
$$
The conserved total amount is $\int_{-\pi}^\pi u\,dx=2\pi^3/3$. 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.
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