 Solution

For uniform unconstrained boundary conditions, the positive gradient coefficient favors a spatially constant minimizer. Put $u=|\mathbf{m}{|}^2\ge 0$. The remaining free-energy difference is $a(T-T_c)u+b{u}^2$. Its minimum isBelow $T_c$ the amplitude is fixed but its direction is arbitrary; the set of uniform minima is a two-sphere. The free-energy minimum is $f_0-a^2(T_c-T{)}^2/(4b)$ below $T_c$ and $f_0$ above. The order parameter vanishes continuously, so this is a continuous, second-order transition within Landau theory.

The free energy has internal $O(3)$ spin-rotation/reflection symmetry; a chosen nonzero vector leaves an $O(2)$ subgroup invariant. Restricting to proper rotations, this is $SO(3)\to SO(2)$ spontaneous symmetry breaking. There are two transverse orientation directions and a massive amplitude direction below the transition. Spatially varying directions cost gradient energy, and imposed boundaries or defects can prevent a uniform minimizer. In strictly two-dimensional equilibrium with short-range interactions, the fluctuation caveat of part (d) still applies: minimizing this polynomial functional at mean-field level does not overturn the absence of finite-temperature spontaneous magnetization.