Under the independence closure, the mean rightward flux across the bond $(i,i+1)$ is
$$
J_{i+1/2}=k_i^+M_i(1-M_{i+1})
-k_{i+1}^-M_{i+1}(1-M_i).
$$
The mean equation is the discrete conservation law $\dot M_i=J_{i-1/2}-J_{i+1/2}$. Substitution of the rates from part c and Taylor expansion show that the exclusion factors cancel from the symmetric diffusive contribution but remain in the biased contribution:
$$
J=-D\partial_xc+v(x)c(1-c).
$$
Therefore the mean-field <asymmetric simple exclusion process> limit is
$$
\boxed{\partial_tc
=D\partial_x^2c-\partial_x[v(x)c(1-c)].}
$$
The factor $1-c$ is the probability that the destination site is vacant.

Solution

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

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The asymmetric simple exclusion process permits at most one particle per lattice site and biases jumps in one direction. A mean-field hydrodynamic limit has advective flux proportional to $c(1-c)$ because both an occupied departure site and vacant arrival site are required.


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