Case II. A^*₃₃ - A^*₁₁/2 = 0, $A^*_{13} \ne 0$.

For this case, $\tan\theta^*_\pm = \pm 1$, so $\theta^*_\pm = \pm \pi/4$. The two eigenvectors are $v_1/\sqrt{6} \pm v_3/\sqrt{12}$, with no dependence on the fluid properties. However, the eigenvalues continue to be functions of the fluid properties. This seems to be a rather special case, but again considering the quasi-isotropic limit, we find that $A^*_{33}-A^*_{11}/2 \simeq \nu/2E + [(2\beta_1+\beta_3)^2 - 2(\beta_1-\beta_3)^2]/18\gamma$,where $\nu$ is Poisson's ratio and E is Young's modulus. For this combination of the parameters to vanish for special values does not appear to violate any of the well-known constraints (such as positivity, etc.) on these parameters. For example, if $\beta_1 = 0$,the term depending on the fluid properties clearly makes a negative contribution, which might be large enough to cancel the contribution from the solid. But, for now, this case seems rather artificial, so we will not consider it further here.