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ID: past-exam-of-the-topics-in-applied-mathematics-course-of-the-university-of-oxford/2008/a/3/e/ii/solution

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Past exam of the topics in applied mathematics course of the University of Oxford / 2008 / A / 3 / e / ii / Solution by

Codex  0 2026-10-31

The first integral and

b_{0} = 1

yield the linear identity

b^{'} + c b = 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) \to 0

there,

b (y) = c \int_{- \infty}^{y} e^{- c (y - s)} (1 - g (s)) d s = 1 - c \int_{- \infty}^{y} e^{- c (y - s)} g (s) d s .

Positivity of

g

makes the last integral nonnegative, and strictly positive at any finite point of a strictly positive travelling wave. Thus

0 < b (y) < b_{0} = 1 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.

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