Past exam of the further mathematical biology course of the University of Oxford / 2023 / 3

The bacteria disperse by diffusion with coefficient $D_B$. Their additional chemotaxis flux isIt points up the nutrient concentration gradient, and its negative spatial derivative enters the balance equation. The positive $N_0$ prevents singular sensitivity at zero nutrient concentration; increasing the background nutrient concentration reduces the drift response to a fixed gradient. The logistic growth equation term $rB(1-B/K)$ represents net reproduction with low-population growth rate $r$ and carrying capacity $K$.

The nutrient undergoes diffusion with coefficient $D_N$. Its production term $a{B}^2/(B_0^2+B^2)$ is an increasing, bounded response to the bacteria, reaching half its maximum $a$ at $B=B_0$. It has the algebraic form of a Hill equation with exponent two; the equation alone does not specify a molecular mechanism. The term $dBN$ removes nutrient at a rate proportional to both bacteria abundance and nutrient concentration.

The full fluxes areThe homogeneous Neumann boundary conditions make both fluxes vanish at the endpoints, so neither bacteria nor nutrient crosses the boundaries. The initial conditions specify their starting spatial distributions. This model includes nutrient production by bacteria, rather than a separate external supply term.

Use $K$ as the population scale, $N_0$ as the nutrient concentration scale, $r^{-1}$ as the time scale, and $L$ as the length scale. The nondimensionalization isFor example, the chemotaxis term becomesDividing the population equation by $rK$ and the nutrient equation by $r{N}_0$ gives the requested system with the following six dimensionless quantities:The rescaled domain is $0<x<1$, $t>0$. Its boundary conditions and initial conditions are

For a spatially uniform steady state with nonzero bacteria abundance, $b(1-b)=0$ forces $b_{\ast }=1$. The nutrient balance then givesTo test spatially uniform linear stability, writeAt the positive equilibrium, the Jacobian matrix isIts eigenvalues are $\boxed{-1,-\mu }$, both negative, so the positive equilibrium is asymptotically stable against spatially uniform perturbations. This remains true at $\mu =1$, where a repeated eigenvalue may produce a decaying $t{e}^{-t}$ contribution. The printed stability claim must be understood in this spatially uniform sense: sufficiently strong chemotaxis can destabilize nonuniform perturbations, as the next answer shows.

Write $b=1+u$, $n=n_{\ast }+v$ and defineThe first-order perturbation equations areThe homogeneous Neumann boundary conditions select Fourier modes proportional to $\cos (m\pi x)$, with squared wavenumber $q_m=(m\pi {)}^2$ for $m=0,1,\dots$. For a mode $e^{\sigma t}(U,V)\cos (m\pi x)$, the eigenvalue problem isIts trace is $-[1+\mu +(d_b+d_n)q]<0$. Therefore a growing mode exists exactly when its determinant is negative. SetThe dispersion relation isThe constant mode has $P(0)=\mu >0$; very short wavelengths also decay because $P(q)\sim d_b{d}_n{q}^2$. For a negative interval at positive $q$, the upward-opening quadratic polynomial must have its minimum below zero at positive $q$. This givesIn particular, $\boxed{b_0>1}$ is necessary, since otherwise $\eta \le 0$ and every coefficient of $P(q)$ is positive. When $b_0>1$, a locally increased population must increase net nutrient production, so attraction to that nutrient reinforces the population increase. This feedback must overcome both diffusion and local relaxation. The calculation is an instance of chemotactic instability with logistic growth and saturating production.

The displayed continuous-wavenumber inequality is necessary on the finite domain, but an allowed mode must actually lie in the unstable band:Equivalently, the exact discrete threshold isEquality gives a marginal zero eigenvalue rather than positive growth. At the unrestricted onset, $q_c=\sqrt{\mu /(d_b{d}_n)}$, so $k_c=(\mu /(d_b{d}_n){)}^{1/4}$. On the finite interval, Discrete Neumann mode selection first destabilizes the mode minimizing the discrete threshold; it need not attain $k_c$ exactly.

To predict the early form of a perturbation, use the larger growth eigenvalueFor $\eta >0$, the square root is real. Choose $m_{\mathrm{fast}}$ maximizing ${\sigma }_{+}(q_m)$ among allowed modes with positive growth. The corresponding perturbation has wavelength $2/m_{\mathrm{fast}}$ in $x$, or $2L/m_{\mathrm{fast}}$ in $X$. Its eigenvector satisfiesso bacteria and nutrient peaks initially coincide. A mode with larger $m$ has finer spatial structure. The two signs of $\cos (m\pi x)$ are both allowed, exchanging peaks and troughs.

The illustration holds $d_b=0.002$, $d_n=0.04$, $\mu =0.5$, $b_0=2$ and $\lambda =2.5$ fixed, giving $n_{\ast }=1$ and $\eta =0.3$, and varies ${\chi }_b$. Crossing the discrete threshold makes a nonuniform Fourier mode grow; stronger chemotaxis widens the unstable band and can favor finer structure. Increasing either diffusion coefficient raises the threshold of each fixed mode. Varying $\lambda$ can have a nonmonotone effect: with $s=1+b_0^2$,is maximal at $\lambda =\mu s$, where $n_{\ast }=1$. Additional production beyond that point weakens sensitivity enough to reduce this feedback. Changes in $L$ also change the allowed dimensional wavenumbers $m\pi /L$ and all three scaled transport coefficients. These are predictions of linear stability analysis: eventual pattern amplitude, competition between growing modes and possible merging require the nonlinear system. The threshold is a necessary condition for spontaneous pattern growth from arbitrarily small perturbations, rather than a claim that every finite-amplitude nonuniform state is excluded below it.