Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 65 / 2 / ii / Solution

Take $A>0$ and a nontrivial nonnegative profile. On $0<r<R(t)$ put $X=(R/r{)}^a$ and $\Sigma =\sigma (X-1)$. The diffusion operator acts on $r^{1/2}\bar{\nu }\Sigma =A{\sigma }^2[r^{5/2-2a}{R}^{2a}-2r^{5/2-a}{R}^a+r^{5/2}]$, givingBut differentiating the trial profile in time gives onlyFor $a\ne 0$, the functions $1,X,X^2$ are independent on the interior. Their coefficients can match only ifMatching the remaining coefficients givesThus $\sigma =[15A(t+t_0){]}^{-1}$, with $R=C(t+t_0{)}^{2/5}$ when $a=1$, and $R=C(t+t_0{)}^{1/2}$ when $a=5/4$. A nonzero solution whose support shrinks to zero at $t=0$ has $t_0=0$. The two compact self-similar disk with density-linear viscosity profiles are consequentlySet $\Sigma =0$ for $r>R$. Both $r^{1/2}\bar{\nu }\Sigma$ and its first radial derivative vanish at the moving edge, because $\bar{\nu }\Sigma$ vanishes quadratically there. Together with $\Sigma (R)=0$, this makes the cutoff consistent with the diffusion equation without an added singular edge flux.

These are singular-origin self-similar solutions: $R(0)=0$ is a limiting support condition, while $\sigma$ diverges as $t\downarrow 0$. It is not ordinary finite pointwise initial data. A finite nonzero $\sigma (0)$ would instead make $R=0$ an invariant solution of the radius equation, precluding an expanding nontrivial profile from that initial radius.