= Solution
For a positive-speed <travelling wave>, the nutrient equation gives $a'=kb/c$. Integrating the bacterial equation once, with zero flux in the uncolonized limit, gives
$$
Db'-\chi b\frac{a'}a=-cb.
$$
Hence
$$
\boxed{b'=\frac{b}{cD}\left(\frac{k\chi b}{a}-c^2\right),\qquad a'=\frac{kb}{c}.}
$$
In the region where $b>0$, divide the two equations. With $\gamma=\chi/D$,
$$
\frac{db}{da}-\frac\gamma a b=-\frac{c^2}{kD}.
$$
The condition $b(1)=0$ fixes the integration constant, giving
$$
\boxed{b(a)=\frac{c^2}{k(\chi-D)}(a-a^{\chi/D})\quad(\chi\ne D).}
$$
At $\chi=D$ 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 $\chi\geq D$ it reaches zero only as $z\to-\infty$.
For $\chi=2D$, the nutrient equation reduces to $a'=(c/D)a(1-a)$ and integration gives
$$
\boxed{a(z)=\frac1{1+Ke^{-cz/D}},\qquad b(z)=\frac{c^2K}{kD}\frac{e^{-cz/D}}{(1+Ke^{-cz/D})^2}.}
$$
The factor $K$ in the bacterial numerator is required by $a'=kb/c$; 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 $\int b\,dz=c/k$. 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.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2009/ii/paper-2-chemotactic-band.png]
{title=Travelling bacterial pulse and nutrient front for logarithmic <chemotaxis> with chi equal to twice D, plotted in dimensionless travelling coordinates}
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