 Solution

For fixed smooth ${\xi }_y,{\xi }_z$, the only appearances of ${\xi }_x$ are in $\delta \Pi =A-Td$ and the nonnegative tension contribution $C{k}_y^2|{\xi }_x{|}^2$. Positivity of $T$ makes the pressure square nonnegative too. Impose $d=A/T$ by choosingThen $\delta \Pi =0$ exactly and ${\xi }_x=O(k_x^{-1})$ as $|k_x|\to \infty$, removing both nonnegative contributions in the limit. Since the remaining terms do not involve ${\xi }_x$, no other choice can lower their limiting value. This is the short transverse wavelength limit for magnetic buoyancy.

Take the chosen vertical and along-field displacements compactly supported in the interior when constructing instability trials; the resulting ${\xi }_x$ also vanishes near the boundaries, satisfying the surface condition. The argument constructs an infimum, not necessarily a minimizing finite-wavelength eigenfunction. Whenever the limiting energy is strictly negative, sufficiently large finite $|k_x|$ already gives negative energy and hence the instability asserted by the supplied variational criterion.