The odd $X_1$ copy number decreases by two until it reaches one, so $\langle x_1(t)\rangle\to1$. The exact $X_2$ first-moment equation is
$$
\dot m_2
=\alpha_2\langle x_1\rangle
-\alpha_3\langle x_2^2\rangle
-\alpha_7\langle x_2x_4\rangle.
$$
Because copy numbers are <nonnegative integers>, $x_2^2\geq x_2$, and the last mixed moment is nonnegative. Hence
$$
\dot m_2+\alpha_3m_2
\leq\alpha_2\langle x_1\rangle.
$$
The <integrating factor> $e^{\alpha_3t}$, together with $\langle x_1(t)\rangle\to1$, gives the comparison bound
$$
\limsup_{t\to\infty}m_2(t)
\leq\frac{\alpha_2}{\alpha_3}.
$$

Subtracting the exact $X_4$ first-moment equation from the $X_3$ equation cancels reaction 5:
$$
\frac d{dt}\left(\langle x_3\rangle-\langle x_4\rangle\right)
=\alpha_4-\alpha_6m_2(t).
$$
The assumed inequality is precisely
$$
\alpha_4-\alpha_6\frac{\alpha_2}{\alpha_3}>0.
$$
Choose a positive $\varepsilon$ smaller than this gap divided by $\alpha_6$. The <limit superior> bound implies that, for all sufficiently large $t$,
$$
\frac d{dt}\left(\langle x_3\rangle-\langle x_4\rangle\right)
\geq
\alpha_4-\alpha_6\left(\frac{\alpha_2}{\alpha_3}+\varepsilon\right)>0.
$$
Thus $\langle x_3\rangle-\langle x_4\rangle$ grows at least linearly. Since $\langle x_4\rangle\geq0$,
$$
\boxed{\lim_{t\to\infty}\langle x_3(t)\rangle=\infty,}
$$
so the required species index is $\boxed{i=3}$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 356 / 1 / e / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-356/1/e/solution