Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 356 / 2 / b

Table of contents
- Solution b

Solution

Multiplying the master equation by

a_{i}

, summing over every state, and shifting indices in each gain term gives, for

2 \leq i \leq m - 1

,

\frac{d M _{i}}{d t} = k_{i - 1}^{+} M_{i - 1} - (k_{i}^{+} + k_{i}^{-}) M_{i} + k_{i + 1}^{-} M_{i + 1} .

At the absorbing left edge and reflecting right edge the corresponding equations are

\frac{d M _{1}}{d t} = - (k_{1}^{+} + k_{1}^{-}) M_{1} + k_{2}^{-} M_{2},

\frac{d M _{m}}{d t} = k_{m - 1}^{+} M_{m - 1} - k_{m}^{-} M_{m} .

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