 Solution

Use complex Fourier amplitudes and let primes denote $z$ derivatives. Put $C=B^2/{\mu }_0$, $T=\gamma p+C$ and $d=i{k}_x{\xi }_x+i{k}_y{\xi }_y+{\xi }_z^{\prime }$. Static balance gives $(p+B^2/(2{\mu }_0){)}^{\prime }=-\rho g$. The perturbations from part (a) then implyFor the last identity, combine $\delta p$ with $B\delta B_y/{\mu }_0$ and use static balance; this is where the background magnetic-pressure gradient cancels.

Multiply the Fourier momentum equation by $-{\xi }^{\ast }$ and integrate from $a$ to $b$. Integrating the pressure-gradient term by parts gives $[{\xi }_z^{\ast }\delta \Pi {]}_a^b-\int d^{\ast }\delta \Pi dz$. The gravity term is $\int [-g{\rho }^{\prime }|{\xi }_z{|}^2-\rho g{\xi }_z^{\ast }d]dz$. The magnetic terms reduce toIndeed the term from $\delta B_z{B}^{\prime }{e}_y$ cancels exactly the $B^{\prime }$ term arising from $i{k}_{y}B\delta B$. This establishes, before completing the square,Now $\delta \Pi =A-Td$, soThus the magnetohydrodynamic energy principle has the requested quadratic formAll squares are squared complex moduli. With the prescribed vanishing surface term the quadratic form is real, as required for the self-adjoint stability problem.