Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2010 / Paper 1 / 2 / 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 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.