Past exam of the mathematics course of the University of Cambridge / 2004 / iii / Paper 65 / 2

Start from the constant-coefficient mean-field dynamo equationHere $\eta$ may include turbulent as well as microscopic diffusion. Use a local Cartesian model, neglect meridional circulation and curvature, and assume independence of $y$. Represent the poloidal magnetic field by a vector potential along $y$ and the toroidal magnetic field by its $y$ component:The shear contributes $(\overline{B}\cdot \nabla )U=\Omega A_x\hat{y}$. The poloidal part of $\alpha \nabla \times (B\hat{y})$ gives $A_t=\alpha B+\eta {\nabla }^{2}A$, while the toroidal part of $\alpha \nabla \times {\overline{B}}_{\mathrm{pol}}$ is $-\alpha {\nabla }^{2}A$. ThereforeRetaining one common transverse dependence $e^{i\ell z}$ changes ${\nabla }^2$ to ${\partial }_x^2-{\ell }^2$, producing the two requested equations. The first coupling is the alpha effect regenerating poloidal field from toroidal field; the shear is differential-rotation winding in an Alpha-Omega dynamo. Retaining the alpha contribution in the second equation also allows alpha-squared dynamo feedback, rather than making the pure Parker shear-dominated approximation.

Include temporal dependence $e^{\sigma t}$ and put $p=k^2+{\ell }^2$. The two Fourier amplitudes obeyConsequently the dispersion relation and its branches areFor $\eta >0$ and $\ell \ne 0$, the branch with positive real square root is the potentially unstable one. At marginal stability set $\sigma =i\omega$; comparing real and imaginary parts givesEliminating $\omega$ produces the alpha-squared Omega dynamo thresholdEither sign of ${\alpha }_m$ is allowed; it changes the wave's propagation sign. Larger $|\alpha |$ increases the real part of the growing branch through zero.

To minimize the threshold differentiate with respect to $x=k^2$. The sign of this derivative is the sign ofIts derivative is strictly positive for $x\ge 0$, so there is at most one interior minimum. Its value at zero changes sign atFor ${\Omega }^2\le {\Omega }_0^2$, $F(x)\ge 0$ and the minimum is $\boxed{k_c=0}$, with $|{\alpha }_c|=\eta |\ell |$. For stronger shear, the minimizing positive wavenumber is the unique rootAn explicit form follows by setting $s={\Omega }^2/(4{\eta }^2{\ell }^4)>1$ and $y=1+k_c^2/{\ell }^2$. The cubic becomes $y^3+3sy-4s=0$, with real solutionThe negative wavenumber $-k_c$ has the same threshold. In the strong-shear limit $k_c\to |\ell |/\sqrt{3}$.

The threshold curve is even in $k$ and behaves as $\eta |k|$ at large $|k|$. Below the shear threshold it increases monotonically away from its central minimum. Above the threshold, zero is a local maximum flanked by the two nonzero minima. At equality the central minimum has vanishing quadratic curvature but remains a minimum. These two cases are sketched from the derived formula below. If $\ell =0$, the zero-wavenumber uniform/gauge mode is degenerate and the positive-wavenumber threshold tends to zero as $k\to 0$; the nonzero-$\ell$ minimum formula must not be used as a nondegenerate onset criterion there.