Past exam of the topics in applied mathematics course of the University of Oxford / 2008 / A / 4

The term $\alpha {\nabla }^{2}q$ is cell diffusion arising from random movement. Chemotaxis up the concentration gradient gives the cell flux $\beta q\nabla c$, whose negative divergence is the transport contribution $-\beta \nabla \cdot (q\nabla c)$. The reaction term $q(1-q)$ gives growth governed by the logistic differential equation with dimensionless carrying capacity one.

The chemoattractant itself diffuses, is produced by cells at rate $\kappa \eta q$, and decays at rate $\kappa c$. The homogeneous Neumann boundary conditions suppress the normal diffusive flux of each quantity. They also suppress the normal chemotactic cell flux because the normal derivative of $c$ vanishes. Thus the rectangular boundary is impermeable, and the spatially integrated cell count changes through cell proliferation rather than transport across that boundary.

For spatially homogeneous perturbations write $q=1+u$ and $c=\eta +v$. The linearized reaction equations areThis triangular matrix has eigenvalues $-1$ and $-\kappa$, both negative. Hence the homogeneous steady state is linearly asymptotically stable against homogeneous perturbations. Even when $\kappa =1$, where the two eigenvalues coincide and a polynomial factor may occur in the matrix exponential, all such perturbations still decay.

For the cosine mode in the question, the Laplacian acts by multiplication by $-s$, whereThese modes obey the Neumann boundary conditions. In the chemotactic term, put $q=1+u$, $c=\eta +v$. Then $\nabla c=\nabla v$, and $q\nabla c=\nabla v+u\nabla v$; the second product is a quadratic polynomial in perturbation amplitude. Thus the linearized equations areSubstituting the modal time dependence givesSeparating this matrix into its transport part and the reaction Jacobian matrix gives the required form, with the same $J$ as above.

For the modal matrix $M(s)$, $s\ge 0$, its trace is always negative:Its determinant isThe proposed inequality is exactly $\beta \kappa \eta <\alpha \kappa +1$. It makes the coefficient of $s$ positive, while the leading and constant coefficients are already positive. Therefore $D(s)>0$ for every $s\ge 0$, independent of which discrete wave numbers the domain allows.

A real two-by-two matrix with negative trace and positive determinant has both eigenvalues with negative real part: if they are real they have positive product and negative sum, and if nonreal their common real part is half the trace. Every mode therefore decays, proving linear stability for all domain sizes under the stated sufficient condition.

Here $L_1=1$ and $L_2=1/2$, so a perturbation varying in both coordinates has $n,m\ge 1$ andWith the given unit coefficients the determinant becomes $D(s)=(1+s{)}^2-\beta s$. The trace remains negative, so instability occurs precisely when $D(s)<0$, orThe right-hand side is strictly increasing for $s>1$, since its derivative is $1-s^{-2}>0$. Its smallest allowed value is attained by the mode $(n,m)=(1,1)$, giving the onset thresholdAt this value the first mixed mode has a zero eigenvalue; it becomes unstable for $\beta >{\beta }_{\mathrm{crit}}$. Thus the critical value is the infimum of the strictly unstable range, rather than a strictly unstable point itself. Its spatial factor is $\cos (\pi x)\cos (2\pi y)$.