Inside the ring, $\nu\Sigma=A\Sigma^3=A\sigma^3(1-x^2/w^2)^{3/2}$. Part a therefore gives
$$
\boxed{u_x=\frac{9A\sigma^2}{w^2}x}.
$$
The edge is material, so $\dot w=u_x(w)$ and
$$
\boxed{\dot w=\frac{9A\sigma^2}{w}}.
$$
Direct substitution of the profile into the diffusion equation gives
$$
\boxed{\dot\sigma=-\frac{\sigma\dot w}{w}},
$$
hence $\sigma w=\sigma_0w_0$. This is exactly <mass conservation>, since
$$
M_{\rm ring}=\int_{-w}^w\Sigma\,dx
=\sigma w\int_{-1}^1\sqrt{1-y^2}\,dy
=\frac\pi2\sigma w.
$$
Using $\sigma=\sigma_0w_0/w$ in the width equation and integrating,
$$
w^4=w_0^4+36A\sigma_0^2w_0^2t.
$$
Therefore
$$
\boxed{
w=w_0\left(1+\frac{36A\sigma_0^2}{w_0^2}t\right)^{1/4},
\qquad
\sigma=\sigma_0\left(1+\frac{36A\sigma_0^2}{w_0^2}t\right)^{-1/4}}.
$$
In particular, $w\propto t^{1/4}$ at late times.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 321 / 2 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-321/2/c/solution