= Solution
The <first integral> and $b_0=1$ yield the linear identity $b'+cb=c(1-g)$. Multiply by the <integrating factor> $e^{cy}$ and integrate from $-\infty$. Since the bounded limit $b(-\infty)=1$ makes $e^{cy}b(y)\to0$ there,
$$
b(y)=c\int_{-\infty}^{y}e^{-c(y-s)}(1-g(s))\,ds
=1-c\int_{-\infty}^{y}e^{-c(y-s)}g(s)\,ds.
$$
Positivity of $g$ makes the last <integral> nonnegative, and strictly positive at any finite point of a strictly positive <travelling wave>. Thus
$$
\boxed{0<b(y)<b_0=1\quad\text{for finite }y,}
$$
with equality approached only at the far-behind boundary. This bound uses the <first integral> rather than an unsupported assumption that the <bacterial> profile is monotone.
Back to article page
