= Solution
For constant diffusivity, the trial <Rayleigh quotient> simplifies to
$$
\nu_{\mathrm{trial}}(q,0)=\frac{(q+1)(q+2)}q=q+3+\frac2q.
$$
Its <derivative> is $1-2/q^2$, so its only positive critical point is $q=\sqrt2$. The second <derivative> is $4/q^3>0$; also the quotient diverges as $q\downarrow0$ or $q\to\infty$. Therefore this is the global optimum within the trial family:
$$
\boxed{q_{\mathrm{opt}}=\sqrt2,\qquad \nu_0\leq\nu_{\mathrm{trial,min}}=3+2\sqrt2\approx5.82843}.
$$
The two requested profiles are
$$
\boxed{T_{\mathrm{trial}}(r,0)=1-r^{\sqrt2},\qquad
T_{\mathrm{trial}}(0,t)=e^{-(3+2\sqrt2)t}}.
$$
The initial radial graph decreases from one at the centre to zero at the rim, has horizontal tangent at the centre, and is concave downward for $0<r<1$. The central temperature decreases exponentially from one to zero, with initial time <derivative> $-(3+2\sqrt2)$. These are trial profiles, not exact solutions of the full <heat equation>. Although the radial trial function is not twice differentiable at the centre, it has finite weighted <energy> and belongs to the variational trial class used by the <Rayleigh-Ritz variational principle>.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2018/paper-1-q10-trial-profiles.png]
{title=Optimal trial radial profile and its assumed exponential decay at the centre}
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