 Solution

Assume $u>0$, as required for the stable quartic model, and retain the stated domain $t>0$. For the opposite-axis quadratic anisotropy of a vector order parameter, fix the magnitude $\rho$. Then the anisotropy lowers the energy most by putting all magnetization along component 1 if $r<0$ and component 2 if $r>0$. Component 3 has positive quadratic coefficient $t$ and cannot lower the uniform minimum. Minimizing the remaining amplitude givesThe component-1 and component-2 ordered regions each break a discrete sign symmetry, giving two easy-axis orientations rather than a sphere of equivalent directions. To check stability, in the component-1 minimum the longitudinal curvature is $-2(t+r)>0$, while the two transverse curvatures are $-2r>0$ and $-r>0$. The component-2 case similarly has longitudinal curvature $-2(t-r)$ and transverse curvatures $2r,r$, all positive in its region. The disordered state is stable exactly when $t\pm r\ge 0$.

Both boundaries $r=-t$ and $r=t$ are continuous, second-order mean-field transitions: the selected component vanishes with a square root, and the first derivatives of the minimum free energy agree across the boundary. In the stipulated upper half-plane $t>0$, the two ordered phases are separated by a disordered wedge. There is no direct first-order boundary between them in that domain. Extrapolating a phase boundary through $t<0$ would answer a different, larger phase-diagram question.