The two transport equations are
$$
\partial_tp_+=-s\partial_xp_+-\lambda_+p_++\lambda_-p_-,
$$
$$
\partial_tp_-=s\partial_xp_-+\lambda_+p_+-\lambda_-p_-.
$$
For $p=p_++p_-$ and $j=s(p_+-p_-)$ they become
$$
p_t=-j_x,
\qquad
j_t=-s^2p_x+2sb c'(x)p-2\lambda_0j.
$$
Eliminating $j$ gives the closed <chemotactic telegraph equation>
$$
\boxed{p_{tt}+2\lambda_0p_t=s^2p_{xx}
-2sb\,\partial_x[c'(x)p].}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 356 / 3 / b / Solution

A chemotactic telegraph equation describes persistent motion with finite turning rate and a turning bias set by a chemical gradient. Its large-turning-rate limit is a drift--diffusion equation with chemotactic drift.


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