Past exam of the mathematics course of the University of Cambridge / 2010 / ii / Paper 3 / 6A (Mathematical Biology)

With no chemotaxis, the stationary oxygen equation is $D_c{c}^{\prime \prime }=k\bar{n}$. Symmetry and the two prescribed surface values giveBy Fick's first law, the vertical oxygen flux is $j_c=-D_c{c}^{\prime }=-k\bar{n}z$. Thus each face supplies an inward flux $k\bar{n}d/2$ per unit area, and the total influx is $k\bar{n}d$, equal to the integrated consumption across the layer. The profile is physically nonnegative only if $c_0\ge k\bar{n}{d}^2/(8D_c)$; otherwise the constant-consumption approximation must be modified.

For bacteria retained in the fluid, impose zero bacterial flux at the faces. The stationary bacterial equation says that $j_n=-D_n{n}^{\prime }+\mu n{c}^{\prime }$ is constant, hence zero. Integrating $n^{\prime }/n=(\mu /D_n)c^{\prime }$ gives $n(z)=n_0\exp (\mu c(z)/D_n)$, with $n_0$ fixed by the total bacterial population. Substitution into the oxygen equation provesZero bacterial through-flux is the boundary assumption behind this equilibrium relation.