= Solution
At a sharp interface, continuity of <velocity> and <traction> gives, for unit incident <velocity> from the left,
$$
1+r_v=t_v,\qquad Z_a(1-r_v)=Z_bt_v.
$$
Thus $r_v=(Z_a-Z_b)/(Z_a+Z_b)$ and $t_v=2Z_a/(Z_a+Z_b)$. Normalizing the transmitted <velocity> by its <elastic-wave energy flux> gives $T_{ba}=\sqrt{Z_b/Z_a}\,t_v$, exactly the result above. Right incidence gives the reversed reflection sign and the same flux-normalized transmission. A <stress> <reflection coefficient> has the opposite sign to the <velocity> <reflection coefficient>; the convention matters.
Put $r=R_{aa}$ and $t=T_{ab}$. The incoming-to-outgoing <matrix> is $S_0=\begin{pmatrix}r&t\\t&-r\end{pmatrix}$, with $r^2+t^2=1$. Therefore \b[$S_0^\dagger S_0=I$], proving equality of incoming and outgoing time-averaged flux for arbitrary coherent incident <wave amplitudes>, not just incidence from one side.
For all real <frequencies> the harmonic equations instead give
$$
\frac{d|\phi_+|^2}{dx}=2g\operatorname{Re}(\phi_+^*\phi_-),\qquad \frac{d|\phi_-|^2}{dx}=2g\operatorname{Re}(\phi_-^*\phi_+).
$$
Their right sides are equal; the diagonal terms $\pm i\omega/\beta$ have zero real contribution. Consequently
$$
\boxed{\frac{d}{dx}\left(|\phi_+|^2-|\phi_-|^2\right)=0.}
$$
The mean flux for real-part phasors is $(|\phi_+|^2-|\phi_-|^2)/2$. Evaluating it at the two endpoints proves exactly that total outgoing flux equals total incoming flux at every real <frequency>. This relies on real lossless moduli and <mass density>; absorption would change the identity.
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