 Solution

Interpret perfect conductivity here as the perfectly conducting thermal boundary condition $\theta =0$; impose the stated magnetic verticality separately. Together with the stress-free boundary condition, admissible vertical profiles areeach multiplied by $e^{i\mathbf{k}\cdot \mathbf{x}+\lambda t}$. Solenoidality requires $i\mathbf{k}\cdot {\mathbf{U}}_h=-\pi W$ and $i\mathbf{k}\cdot {\mathbf{B}}_h=\pi Z$. The resistive induction equation preserves both profile choices. In particular ${\mathbf{b}}_h=0$ at each boundary and ${\partial }_z{b}_z=0$ there.

Set $s=k^2+{\pi }^2$ and $p={\pi }^2$. The pressure-projected vertical momentum, thermal and vertical induction equations reduce toTheir determinant gives the vertical-field magnetoconvection dispersion relationIt is best kept as a polynomial rather than canceling a factor that might vanish at a special eigenvalue. For $\lambda =0$,For an oscillatory marginal mode $\lambda =i\omega$ with $\omega \ne 0$, write the polynomial as ${\lambda }^3+a_1{\lambda }^2+a_2\lambda +a_3$. Its coefficients areSeparating real and imaginary parts gives $a_3=a_1{a}_2$ and ${\omega }^2=a_2>0$. Solving yieldsTherefore oscillatory marginality in vertical-field magnetoconvection requiresThe equality has zero frequency and is not an oscillatory marginal mode. For some positive horizontal wavenumber to satisfy this condition, it is necessary and sufficient that $Q>\zeta (1+\sigma ){\pi }^2/[\sigma (1-\zeta )]$, although the associated threshold must still be minimized over admissible modes. When ${\omega }^2>0$, $R^{(o)}<R^{(e)}$ at the same wavenumber; this also follows from $R^{(e)}-R^{(o)}=(\sigma +1+\zeta )s{\omega }^2/(\sigma \zeta k^2)$.