Past exam of the mathematics course of the University of Cambridge / 2009 / ii / Paper 2 / 13A / Solution

For a positive-speed travelling wave, the nutrient equation gives $a^{\prime }=kb/c$. Integrating the bacterial equation once, with zero flux in the uncolonized limit, givesHenceIn the region where $b>0$, divide the two equations. With $\gamma =\chi /D$,The condition $b(1)=0$ fixes the integration constant, givingAt $\chi =D$ the limiting expression is $b=-(c^2/kD)a\log a$. These formulas are positive between zero and one. For $\chi <D$ the nutrient reaches zero at a finite rear coordinate, so the strictly positive smooth whole-line ansatz with logarithmic chemotactic sensitivity needs a separate zero-nutrient interpretation there; for $\chi \ge D$ it reaches zero only as $z\to -\infty$.

For $\chi =2D$, the nutrient equation reduces to $a^{\prime }=(c/D)a(1-a)$ and integration givesThe factor $K$ in the bacterial numerator is required by $a^{\prime }=kb/c$; it is absent in the printed expression unless $K=1$. A translation of $z$ sets $K=1$, which is the version plotted below. The nutrient rises monotonically from zero behind the band to one ahead, while bacteria form a localized pulse with maximum $c^2/(4kD)$ at the nutrient half-height. Its total population is $\int bdz=c/k$. The band advances into fresh nutrient, consumes it and leaves a depleted region behind; chemotaxis draws bacteria up the nutrient gradient while diffusion spreads them.