Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 68 / 3

Take the coupling parameters to be real. The dipole and quadrupole parity in a mean-field dynamo subspaces are invariant: substitution of $A_2=A_1$, $B_2=-B_1$ makes the second-hemisphere equations equal to the first-hemisphere equations with these same signs. Substitution of $A_2=-A_1$, $B_2=B_1$ likewise preserves the other parity. In either subspace put $A=A_1$, $B=B_1$. The reduced equations and characteristic equations areA marginal normal mode has $\lambda =i\omega$. Equating real and imaginary parts givesConsequently the marginal surfaces areTo identify which side actually grows, write $z=ab$ in $(\lambda +1{)}^2=z$. The largest real part of the two eigenvalues is $-1+\sqrt{(|z|+\mathrm{Re}z)/2}$. It is negative exactly when $\mathrm{Re}z+[\mathrm{Im}z{]}^2/4<1$: squaring $|z|<2-\mathrm{Re}z$ proves this, and the latter condition ensures the right-hand side is positive. Thus dipole modes decay for $F_d<1$ and quadrupole modes decay for $F_q<1$, with growth for the corresponding reversed inequality.

For explicit coupled-hemisphere dynamo parity thresholds, suppose $\Omega \ne 0$ and defineThe dipole stability interval is $\delta -L<{\alpha }_0<\delta +L$, and the quadrupole stability interval is $-\delta -L<{\alpha }_0<-\delta +L$. A comparison of first onset from ${\alpha }_0=0$ assumes both are initially stable, namelyIncrease $|{\alpha }_0|$ along a specified sign $s=\mathrm{sgn}{\alpha }_0$. The first thresholds for the two parities areTherefore the quadrupole reaches onset first if $s({\Omega }^2{ϵ}_1+2{ϵ}_2)>0$; the dipole reaches onset first if it is negative; the thresholds coincide if it vanishes. For positive ${\alpha }_0$ and positive coupling parameters this selects the quadrupole, provided the initial stability condition holds. Reversing the sign of ${\alpha }_0$ exchanges the preference. If both signs are freely allowed, each parity attains the same minimum threshold $|\Omega |(L-|\delta |)$ on opposite sign rays; a unique preference cannot be inferred from the absolute product alone. If the initial stability condition fails, the zero-alpha state is already unstable or marginal and there is no first onset from a stable origin. When $\Omega =0$, the formulas for $F_d,F_q$ still apply, but $|{\alpha }_0\Omega |$ is identically zero and is not a useful onset parameter.

For an alpha-quenching rotating wave, write the four complex fields as constant amplitudes times $e^{i\omega t}$, with $\omega$ real. The squared toroidal amplitudes are then constant, so the algebraic alpha quenching coefficients are constant on the orbit. Assume $\Omega \ne 0$. With ${ϵ}_2=0$, 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 $1-{\omega }^2=0$. Thus every nonzero rotating wave of this form hasSet $a={\alpha }_0-2\omega /\Omega$. In the dipole subspace $B_2=-B_1$ and $A_2=A_1$; in the quadrupole subspace $B_2=B_1$ and $A_2=-A_1$. The amplitude equations giveThese are nonzero waves exactly when the respective squared amplitude is positive. A common constant phase of $B_1,B_2$ is arbitrary, and $|A_j{|}^2=2|B_j{|}^2/{\Omega }^2$. In particular, choosing $\omega =\mathrm{sgn}\Omega$, both parities exist once ${\alpha }_0>2/|\Omega |+|{ϵ}_1|$. 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 $p|B{|}^2$ independently of the sign of ${\alpha }_0$. Thus a sufficiently large positive ${\alpha }_0$ gives the intended waves, but a sufficiently large negative ${\alpha }_0$ makes every displayed squared amplitude negative for both frequency signs. For example, with $\Omega =1$, ${ϵ}_1=0$, ${\alpha }_0=-M$ and $M>2$, neither $(-M-2)/p$ nor $(-M+2)/p$ is nonnegative. Large $|{\alpha }_0\Omega |$ 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 ${ϵ}_1\ne 0$. 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 $B_1=b_1$, $B_2=b_2$ real. The amplitude equations reduce toMultiplying the first equation by $b_2$, the second by $b_1$, and subtracting givesFor unequal magnitudes, therefore, $b_1{b}_2=-{ϵ}_1/p$. Neither amplitude can vanish. Multiplying the first amplitude equation by $b_1$ then shows $ap{b}_1^2-p^2{b}_1^4-{ϵ}_1^2=0$. Equivalently,The two distinct positive roots exist precisely when $a>2|{ϵ}_1|$. This constructs the desired solutions explicitly:Choose $b_1$ to be either square root of the first value, and set $b_2=-{ϵ}_1/(p{b}_1)$; substituting verifies both amplitude equations. Interchanging the two roots gives the hemisphere-exchanged companion. The hinted choice $\omega =1$ therefore requires ${\alpha }_0-2/\Omega >2|{ϵ}_1|$. At equality the magnitudes coincide, so it is not an asymmetric solution. For ${ϵ}_1>0$ the amplitudes have opposite signs and meet the dipole branch at equality; for ${ϵ}_1<0$ they have the same sign and meet the quadrupole branch.

If ${ϵ}_1=0$, the hemispheres decouple. An asymmetric rotating wave instead has one hemisphere inactive and the other with $|B{|}^2=a/p>0$. If both hemispheres are active their magnitudes must be equal, although their relative phase is arbitrary. If $\Omega =0$, the toroidal equations simply give ${\dot{B}}_j=-B_j$, so no nonzero common-frequency rotating wave exists.

The unequal-amplitude solutions describe a periodically reversing dynamo concentrated more strongly in one hemisphere, with $|A_1|\ne |A_2|$ following from $|A_j|=\sqrt{2}|B_j|/|\Omega |$. Both $B_d=(B_1-B_2)/2$ and $B_q=(B_1+B_2)/2$ 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.