Past exam of the mathematics course of the University of Cambridge 2007 iii Paper 68 3 ii Solution Created 2026-10-03 Updated 2026-10-14
For an alpha-quenching rotating wave, write the four complex fields as constant amplitudes times , with real. The squared toroidal amplitudes are then constant, so the algebraic alpha quenching coefficients are constant on the orbit. Assume . With , the toroidal equations determineSubstitution in the potential equations gives the amplitude eigenvalue problemThe matrix is real symmetric, hence Hermitian, and every one of its eigenvalues is real. For a nonzero amplitude vector this forces . Thus every nonzero rotating wave of this form hasSet . In the dipole subspace and ; in the quadrupole subspace and . The amplitude equations giveThese are nonzero waves exactly when the respective squared amplitude is positive. A common constant phase of is arbitrary, and . In particular, choosing , both parities exist once . The other frequency sign is also algebraically possible whenever its positivity conditions hold; existence alone says nothing about stability of that branch.
There is a genuine sign qualification in the polynomial alpha quenching law. It subtracts independently of the sign of . Thus a sufficiently large positive gives the intended waves, but a sufficiently large negative makes every displayed squared amplitude negative for both frequency signs. For example, with , , and , neither nor is nonnegative. Large alone is insufficient as printed; positive alpha, or the exact amplitude-positivity conditions above, is required. A sign-preserving quenching prescription would be a different model.
For the unequal-hemisphere dynamo rotating wave, first take . The real symmetric matrix above has distinct eigenvalues because its off-diagonal entry is nonzero. Its eigenvector is real up to a common complex phase, so choose , real. The amplitude equations reduce toMultiplying the first equation by , the second by , and subtracting givesFor unequal magnitudes, therefore, . Neither amplitude can vanish. Multiplying the first amplitude equation by then shows . Equivalently,The two distinct positive roots exist precisely when . This constructs the desired solutions explicitly:Choose to be either square root of the first value, and set ; substituting verifies both amplitude equations. Interchanging the two roots gives the hemisphere-exchanged companion. The hinted choice therefore requires . At equality the magnitudes coincide, so it is not an asymmetric solution. For the amplitudes have opposite signs and meet the dipole branch at equality; for they have the same sign and meet the quadrupole branch.
If , the hemispheres decouple. An asymmetric rotating wave instead has one hemisphere inactive and the other with . If both hemispheres are active their magnitudes must be equal, although their relative phase is arbitrary. If , the toroidal equations simply give , so no nonzero common-frequency rotating wave exists.
The unequal-amplitude solutions describe a periodically reversing dynamo concentrated more strongly in one hemisphere, with following from . Both and are nonzero, so the field combines the two dipole and quadrupole parity in a mean-field dynamo classes rather than belonging to either one. Equatorial reflection exchanges the two companion solutions. The model can therefore break hemispheric symmetry without asymmetric input parameters. This construction proves existence; nonlinear stability would require an additional perturbation calculation.