Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2013/iii/paper-66/2/solution

Use the homogeneous equilibrium point condition . Evaluate at and abbreviate
The quantity is the chemical relaxation rate with the cell density held fixed. Derivatives of do not enter this linear stability analysis: they multiply spatial derivatives of the homogeneous background or products of perturbations. For a Fourier mode with time dependence , put and write
Consequently
At , the eigenvalues are and . The neutral cell-density mode reflects mass conservation, not decay of every homogeneous perturbation.
Write , , , and interpret production physically as . If , the trace-determinant stability criterion shows that an unstable nonzero wavenumber exists precisely when . The unstable band is
If , the chemical field is already unstable at zero wavenumber; arbitrarily small positive wavenumbers are unstable too. Since , this also implies . Thus, on the infinite plane, the undivided Keller--Segel aggregation threshold is
For , dividing by gives the requested form
At equality there is no strictly growing mode when ; nonzero spatial modes decay. On a finite domain, a permitted nonzero wavenumber must actually lie in the unstable band. If one allows a signed production function, the displayed matrix and trace-determinant stability criterion remain valid, but the simplification using must be revisited.
The printed hypotheses do not ensure . Positivity of the degradation rate alone is insufficient. For a concrete counterexample, take , , , and . The homogeneous equilibrium point condition holds, all rates and transport coefficients are positive, but . The printed left-hand side equals , although for . These spatial modes grow. If , the printed expression is undefined. The undivided criterion and the matrix above resolve both cases.
To find the fastest-growing Keller--Segel mode, use the larger eigenvalue
For the square root is real. Put and . Differentiation gives
A positive maximizing wavenumber exists in either of two growing cases: , , or , . Equivalently, . In this range , and
Thus the stationary point is the unique maximum. The derivative condition yields
Solving it, with the root that satisfies the unsquared derivative equation, gives
For , the unequal-diffusivity expression simplifies to , also agreeing with the equal-diffusivity limit. The formula maximizes the full two-field growth rate; it makes no instantaneous-chemical approximation. Every direction of with this length is equivalent by rotational symmetry.
If but , the maximum instead occurs at : the fastest mode is homogeneous, with and infinite wavelength. For , its initial derivative is . If the derivative thereafter decreases, while if it increases towards the still-negative large- limit; either way no positive- maximum is missed. At , the two eigenvalues are simply and , giving the same conclusion. In a finite box, maximize over the allowed Fourier modes; excluding the homogeneous mode can change the selected length. In a stable parameter range there is no fastest-growing mode.
The first ratio compares the positive feedback loop “more cells produce more attractant, which draws in more cells” with spreading by cell diffusion and removal of chemical perturbations. Its numerator measures chemotaxis together with attractant production; its denominator measures dispersal together with incremental degradation. This competition between directed chemotaxis and cell diffusion produces spatial aggregation. The second ratio compares the concentration dependence of chemical production, , with incremental degradation . Positive amplifies chemical fluctuations, whereas negative suppresses them. It changes the chemical relaxation available to the chemotaxis feedback loop and can itself destabilize the homogeneous chemical field. Chemical diffusion suppresses short scales and sets the selected wavelength, but does not change the infinite-plane long-wave threshold. These interpretations as ratios of stabilizing and destabilizing processes assume .

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