Paper 2, Section II, A

Mathematical Biology | Part II, 2009

Travelling bands of microorganisms, chemotactically directed, move into a food source, consuming it as they go. A model for this is given by

bt=∂∂x[Dbx−bχaax],at=−kbb_{t}=\frac{\partial}{\partial x}\left[D b_{x}-\frac{b \chi}{a} a_{x}\right], \quad a_{t}=-k b

where b(x,t)b(x, t) and a(x,t)a(x, t) are the bacteria and nutrient respectively and D,χD, \chi, and kk are positive constants. Look for travelling wave solutions, as functions of z=x−ctz=x-c t where cc is the wave speed, with the boundary conditions b→0b \rightarrow 0 as ∣z∣→∞,a→0|z| \rightarrow \infty, a \rightarrow 0 as z→−∞z \rightarrow-\infty, a→1a \rightarrow 1 as z→∞z \rightarrow \infty. Hence show that b(z)b(z) and a(z)a(z) satisfy

b′=bcD[kbχa−c2],a′=kbcb^{\prime}=\frac{b}{c D}\left[\frac{k b \chi}{a}-c^{2}\right], \quad a^{\prime}=\frac{k b}{c}

where the prime denotes differentiation with respect to zz. Integrating db/dad b / d a, find an algebraic relationship between b(z)b(z) and a(z)a(z).

In the special case where χ=2D\chi=2 D show that

a(z)=[1+Ke−cz/D]−1,b(z)=c2kDe−cz/D[1+Ke−cz/D]−2a(z)=\left[1+K e^{-c z / D}\right]^{-1}, \quad b(z)=\frac{c^{2}}{k D} e^{-c z / D}\left[1+K e^{-c z / D}\right]^{-2}

where KK is an arbitrary positive constant which is equivalent to a linear translation; it may be set to 1 . Sketch the wave solutions and explain the biological interpretation.

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