Use <stochastic chemical kinetics>[stochastic mass-action propensities]
$$
a_1=\frac{\alpha_1}{V}y(y-1),
\quad a_2=\frac{\alpha_2}{V}x(x-1),
\quad a_3=\frac{\alpha_3V}{\varepsilon},
\quad a_4=\frac{\alpha_4}{\varepsilon V}xy.
$$
Writing $E_x^rf(x,y)=f(x+r,y)$ and similarly for $E_y$, the fast and slow forward operators are
$$
\boxed{\mathcal L_0^*
=(E_y^{-1}-1)\alpha_3V
+(E_y^{+1}-1)\frac{\alpha_4xy}{V},}
$$
$$
\boxed{\mathcal L_1^*
=(E_x^{-1}E_y^{+2}-1)\frac{\alpha_1y(y-1)}V
+(E_x^{+1}-1)\frac{\alpha_2x(x-1)}V.}
$$
Each shift operator acts on everything to its right, including the propensity. Birth of $Y$ and consumption of $Y$ by $X+Y\to X$ are fast; $2Y\to X$ and $2X\to X$ are slow. The slow species is $X$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 356 / 3 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-356/3/a/solution