Solution
ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/ii/paper-2/5e/solution
Past exam of the mathematics course of the University of Cambridge 2015 ii Paper 2 5E Solution by
Codex 0 Created 2026-09-24 Updated 2026-10-06
At a positive spatially homogeneous equilibrium point, and . Hence . The reaction Jacobian matrix there isThe two-dimensional Routh-Hurwitz stability criterion gives strict linear asymptotic stability for , and instability for .
For a spatial mode with , the reaction-diffusion linearization is . When , its trace is negative for every , whileA growing mode exists precisely when this quadratic is negative at some . Its minimum must occur at positive , and must be strictly negative, givingWith , the latter inequality becomes . Selecting the root compatible with yields the Turing instability thresholdThe unstable band isOn a bounded domain a boundary condition must permit a mode in this band. Equality at the threshold gives a neutral mode, not positive growth.
At , the homogeneous linearization has purely imaginary eigenvalues, so it is not strictly linearly asymptotically stable. If “stable” is read as nonlinear stability, the boundary deserves attention. Put , , and , . The exact equations give , , where ,Expanding through cubic order, cancels the quadratic displacement. The correction from the quadratic change in also integrates to zero, because it is proportional to . Since , the Poincaré return map isThis cubic return-map stability of a weak focus shows that is a weak attracting focus. The strict linear range used in the Turing instability calculation is ; the positive homogeneous equilibrium is also nonlinearly asymptotically stable at the boundary . For that boundary, for every nonzero spatial mode. The same strict Turing instability threshold and unstable band therefore also apply there, with the homogeneous mode understood through nonlinear stability.
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