Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 60 / 2 / Solution

The azimuthal derivative of a cylindrical vector Fourier mode must include basis rotation: the azimuthal dependence refers to the components in the rotating cylindrical basis. Since ${\partial }_{\varphi }\hat{\mathbf{s}}=\hat{\mathbf{\varphi }}$ and ${\partial }_{\varphi }\hat{\mathbf{\varphi }}=-\hat{\mathbf{s}}$, a vector amplitude $\mathbf{v}\propto e^{im\varphi }$ obeysAlso $\mathbf{B}=(J/2)(-y,x,0)$ in Cartesian coordinates, so $(\mathbf{v}\cdot \nabla )\mathbf{B}=(J/2)\hat{\mathbf{z}}\times \mathbf{v}$. Consequentlyand the basis-rotation terms cancel in the ideal magnetohydrodynamic induction equation:Here the printed $J$ is the coefficient in the specified magnetic field: $\nabla \times \mathbf{B}=J\hat{\mathbf{z}}$. If $J_{\mathrm{phys}}$ denotes SI electric current density, then $J={\mu }_0{J}_{\mathrm{phys}}$. The field and all subsequent printed coefficients are mutually consistent with this normalization.

For a normal mode with $\sigma \ne 0$, induction gives $\mathbf{b}=imJ\mathbf{u}/(2\sigma )$. Substitution into the momentum equation, including the Coriolis force, gives the reduced velocity equationThus the coefficients are the printed $\lambda$ and $\nu$. Let $C=J^2/({\mu }_0\rho )$. Inserting the allowed relation $\lambda =i\delta \nu$ and multiplying by $\sigma$ yieldsHence the uniform-current rotating magnetohydrodynamic wave dispersion isThe radicand is real. For $|m|\ge 2$, $2m\delta -m^2<0$ because $|\delta |<1$, so both roots are purely imaginary. For $m=0$ the roots of the quadratic are also imaginary; any zero-frequency case must be checked in the original equations, since the elimination divided by $\sigma$. A positive real part requiresFor integer $m$ this is possible only when $|m|=1$ and $m\delta >1/2$. Taking the conventional representative $m\ge 0$ gives the printed exception $m=1$, with $\delta >1/2$ and sufficiently large $J$. Literally allowing negative integers also gives $m=-1$, $\delta <-1/2$; complex conjugation maps $(m,\delta ,\sigma )$ to $(-m,-\delta ,\overline{\sigma })$, so these describe the conjugate real disturbance. Purely imaginary roots describe neutral oscillatory modes; a repeated root at threshold does not itself prove boundedness of arbitrary initial perturbations.