Let $w_{ij}$ be the transition rate from $i$ to $j$ and define the probability current $J_{ij}=w_{ij}P_i-w_{ji}P_j=-J_{ji}$. The <master equation> is $\dot P_i=-\sum_jJ_{ij}$. Differentiating the Shannon entropy $S_{\rm sys}=-\sum_iP_i\log P_i$ and symmetrizing gives
$$
\dot S_{\rm sys}=\frac12\sum_{i,j}J_{ij}\log\frac{P_i}{P_j}.
$$
Adding the entropy flow to the environment produces the nonnegative <entropy production rate of a Markov chain>
$$
\boxed{\dot S_{\rm tot}=\frac12\sum_{i,j}J_{ij}
\log\frac{w_{ij}P_i}{w_{ji}P_j}\geq0.}
$$
The paper's displayed $\sum P\log P$ is the negative of the thermodynamic system entropy, so its derivative has the opposite sign before the environmental contribution is added.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 356 / 1 / a / Solution

For edge current $J_{ij}=w_{ij}P_i-w_{ji}P_j$, the total entropy production rate is
$$
\dot S_{\rm tot}=\frac12\sum_{i,j}J_{ij}
\log\frac{w_{ij}P_i}{w_{ji}P_j}\geq0.
$$
It vanishes precisely when every occupied edge satisfies detailed balance.


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