Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 337 1 iii Solution Created 2026-10-03 Updated 2026-10-06
Eliminate the instantaneous Stokes flow velocity in favour of temperature. On a horizontal Fourier mode , the Stokes temperature-slaving operator maps to , where and . The temperature evolution has linear operator and bilinear map . Under the homogeneous thermal Dirichlet boundary conditions, is self-adjoint. Normalize its critical eigenfunction as and set ; the critical vertical velocity is .
At order , the critical eigenfunction equation gives . At order , the weakly nonlinear expansion contains the imposed second harmonic and the quadratic products of the critical mode: a horizontally uniform temperature correction proportional to and, in a general vertical-mode calculation, a second harmonic proportional to . These corrections are found by solving the noncritical boundary value problems, with homogeneous thermal data except for the imposed forcing.
At order , the method of multiple scales produces the slow derivative , the detuning term , and the two cross-advection terms involving first- and second-order fields. Project the component onto the adjoint eigenfunction using the vertical inner product. This is the solvability condition in the method of multiple scales: divide each resonant projection by . The detuning supplies with ; interactions of horizontal wavenumbers and permit with ; self-interaction through the slaved mean and second harmonic supplies . Other products have the wrong horizontal wavenumber. Reflection permits real coefficients with this cosine forcing. Thus the symmetry-allowed spatially forced convection amplitude equation isThere is a useful specialization that should not be silently missed. For the literal one-vertical-mode Stokes flow problem, the vanishing two-to-one forcing coefficient for Stokes convection makes at this order. To see this, write a positive second-harmonic forcing component as , incorporating the cosine's factor . Its coupling to the negative critical harmonic has projected integrand, apart from sign and its factor ,The integral vanishes because at both plates, even though is nonzero. This proves the cancellation without solving the forced profiles. The permitted coefficient is therefore zero times ; symmetry alone does not establish nonzero phase pinning for the equations actually supplied.
The same normalization makes the remaining coefficients explicit. Since , . The quadratic second harmonic cancels for , while the uniform correction is . Projecting gives . Thus for the literal model and this temperature normalization,A generic nonzero would require a nonvanishing projection in an amended physical model or a different forcing structure. It is still meaningful to classify the real-coefficient amplitude equation requested independently.
Write . Then and . These are a gradient flow for , so local minima give stable equilibrium points. At the origin the two eigenvalues are and . The origin has exponential asymptotic stability if , retains asymptotic stability with algebraic decay at , and is unstable if . At equality, obeys , since both linear coefficients are nonpositive. Integrating this inequality proves attraction even in the zero-eigenvalue direction.
For the stable nonzero equilibrium points are real; for they are imaginary:The real branch has Jacobian matrix eigenvalues ; the imaginary branch has . The oppositely aligned branch, when it exists, is a saddle equilibrium. No mixed real-imaginary nonzero equilibrium is possible when .
For , the origin is stable for , with algebraic decay at zero. If , the circle is radially attracting. Each point has Lyapunov stability but has a neutral phase direction, so it does not have individual asymptotic stability; the circle has orbital stability. This is the literal model's unpinned family. The general nonzero- branches instead exhibit phase locking to one of two phases separated by .