Choose vertices $x,y$ at distance $d_G$, put $r=\lfloor(d_G-1)/4\rfloor$ and $R=\lfloor(d_G-1)/2\rfloor$. The radius-$R$ balls about $x$ and $y$ are disjoint, so one has stationary mass at most $1/2$; call its center $z$. Apply part b from radius $r$ to radius $R$:
$$
\frac12\geq\pi^G(B(z,R))
\geq\pi_*^G(1+\Phi_*^G)^{R-r}.
$$
Here the exponent should be $R-r$, and $R-r\geq(d_G-2)/4$. Therefore
$$
\Phi_*^G
\leq(2\pi_*^G)^{-4/(d_G-2)}-1.
$$
Since the <relaxation time> is $t_{\rm rel}^G=1/\gamma^G$, part a yields
$$
t_{\rm rel}^G
\geq
\frac1{2\Phi_*^G}
\geq
\frac1{2\{(2\pi_*^G)^{-4/(d_G-2)}-1\}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 215 / 4 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-215/4/c/solution