= Solution
Define $K=Z_+-Z_-$. Its antisymmetric off-diagonal entries cancel, leaving
$$
\boxed{K=\frac{2\rho\omega}{D}\operatorname{diag}(p,s),\qquad K^{-1}=\frac{D}{2\rho\omega}\operatorname{diag}(1/p,1/s).}
$$
From $\mathbf v=\mathbf v^++\mathbf v^-$ and $\mathbf t=-Z_+\mathbf v^+-Z_-\mathbf v^-$, solve explicitly for the <P-SV characteristic reconstruction>:
$$
\boxed{\mathbf v^+=-K^{-1}(\mathbf t+Z_-\mathbf v),\qquad\mathbf v^-=K^{-1}(\mathbf t+Z_+\mathbf v).}
$$
The prescribed strictly positive $p,s$ make this decomposition unique; a grazing or <evanescent> branch is outside these assumptions.
In the mean flux, the two mixed directional terms are
$$
\frac12\operatorname{Re}\left[(\mathbf v^-)^\dagger Z_+\mathbf v^++(\mathbf v^+)^\dagger Z_-\mathbf v^-\right]=0,
$$
because $Z_-=-Z_+^\dagger$ makes the second scalar minus the <complex conjugate> of the first. The remaining self terms give
$$
\boxed{\langle S_z\rangle=\frac{\rho\omega}{2D}\left[p\left(|v_x^+|^2-|v_x^-|^2\right)+s\left(|v_z^+|^2-|v_z^-|^2\right)\right].}
$$
Finally use either directional velocity-potential <matrix>. Its weighted quadratic form satisfies
$$
p|v_x^\pm|^2+s|v_z^\pm|^2=\omega^2D\left(p|\phi_\pm|^2+s|\psi_\pm|^2\right).
$$
Indeed the two P-SV cross terms have coefficients $-kps$ and $+kps$ for the forward field, with the signs reversed for the backward field, so they cancel. Substitution proves
$$
\boxed{\langle S_z\rangle=\frac{\rho\omega^3}{2}\left[p\left(|\phi_+|^2-|\phi_-|^2\right)+s\left(|\psi_+|^2-|\psi_-|^2\right)\right].}
$$
Thus net normal <elastic-wave energy flux> is the difference of the independently flux-weighted P and SV directional powers, even for coherent mixtures.
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