Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 356 / 1 / d / i

Solution

f

\frac{d}{d t} ⟨ f ⟩ = ⟨ L f ⟩ .

m (t) = ⟨ x_{2} ⟩

and

s (t) = ⟨ x_{2}^{2} ⟩

. With

α_{7} = 0

, reaction 2 changes

x_{2}

+ 1

and reaction 3 changes it by

- 1

. Therefore

\overset{m}{˙} = α_{2} ⟨ x_{1} ⟩ - α_{3} ⟨ x_{2}^{2} ⟩

and, using

(x_{2} + 1)^{2} - x_{2}^{2} = 2 x_{2} + 1

and

(x_{2} - 1)^{2} - x_{2}^{2} = - 2 x_{2} + 1

\overset{s}{˙} = α_{2} (2 ⟨ x_{1} x_{2} ⟩ + ⟨ x_{1} ⟩) - 2 α_{3} ⟨ x_{2}^{3} ⟩ + α_{3} ⟨ x_{2}^{2} ⟩ .

The equation for the second moment contains the third, whose equation contains the fourth, and so on. This is an infinite moment hierarchy.

Reaction 1 eventually leaves the odd initial copy number at

x_{1}^{*} = 1

. Put

μ = ⟨ x_{2}^{*} ⟩, q = \frac{α _{2}}{α _{3}} .

The stationary first-moment equation gives

⟨(x_{2}^{*})^{2} ⟩ = q .

The stationary second-moment equation then gives

⟨(x_{2}^{*})^{3} ⟩ = q (μ + 1) .

0 = ⟨ (x_{2}^{*} - μ)^{3} ⟩ = ⟨(x_{2}^{*})^{3} ⟩ - 3 μ ⟨(x_{2}^{*})^{2} ⟩ + 2 μ^{3} .

Substitution yields the requested polynomial equation

2 μ^{3} - 2 \frac{α _{2}}{α _{3}} μ + \frac{α _{2}}{α _{3}} = 0.

This cubic comes from the closure approximation; it is not an exact equation for the stationary mean.

There are no discussions about this article yet.

There are currently no matching articles.