Past exam of the topics in applied mathematics course of the University of Oxford / 2011 / A / 3 / b / Solution

Choose any positive chemical concentration scale

A_{0}

and set

x = L x, t = t / r, N = K n, A = A_{0} a .

After dropping hats, substitution gives

n_{t} = D_{n} n_{xx} - χ (n a_{x})_{x} + n (1 - n), a_{t} = D_{a} a_{xx} + α n - β a,

where

D_{n} = \frac{D _{N}}{r L ^{2}}, D_{a} = \frac{D _{A}}{r L ^{2}}, χ = \frac{χ ˉ A _{0}}{r L ^{2}}, α = \frac{σ K}{r A _{0}}, β = \frac{μ}{r} .

For explicit dimensional checks, write bacterial-density and chemical-concentration units as

N, A

. Then

[D_{N}] = [D_{A}] = L^{2} / T

,

[r] = [μ] = T^{- 1}

,

[\overset{χ}{ˉ}] = L^{2} / (T A)

and

[σ] = A / (T N)

. Since

[K] = N

and

[A_{0}] = A

, each displayed ratio has unit one. The new spatial domain is

[0, 1]

.

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