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Take shear-horizontal waves with displacement $\mathbf{u}=(0,0,w(x,y,t))$. This is divergence-free, so the elastic equation reduces to $\rho w_{tt}=\mu (w_{xx}+w_{yy})$. Rigid boundaries require $w=0$ at $y=0,h$, giving modes $w=A\sin (n\pi y/h)\cos (kx-\omega t)$, $n\ge 1$. With shear speed $c_s=\sqrt{\mu /\rho }$, the dispersion relation isFor $k>0$, the phase velocity is $v_p=\omega /k>c_s$ and the group velocity is $v_g=d\omega /dk=c_s^{2}k/\omega <c_s$, with $v_p{v}_g=c_s^2$.

The kinetic energy density is $T=\rho w_t^2/2$. The only strains are $e_{xz}=w_x/2$ and $e_{yz}=w_y/2$, together with their symmetric counterparts; thus the isotropic linear-elastic energy density is $W=\mu (w_x^2+w_y^2)/2$. Average over one temporal period and integrate across the layer. The sine- and cosine-squared transverse integrals are both $h/2$, givingThe guided shear-horizontal mode has average equipartition after both operations; a time average at a fixed transverse point alone need not give equality.

Use the real modeThe kinetic energy density and isotropic linear-elastic energy density areAverage over one period and integrate across the layer. Since ${\int }_0^h{\sin }^2(q_{n}y)dy={\int }_0^h{\cos }^2(q_{n}y)dy=h/2$, the mean energy per unit horizontal area isThe $x$-directed elastic-energy flux is $S_x=-{\sigma }_{zx}{w}_t=-\mu w_x{w}_t$. Its corresponding average isUsing ${\omega }^2=c_S^2(k^2+q_n^2)$ givesand therefore

For the displacement gradient tensor specified in the question,while the divergence is $\nabla \cdot \mathbf{u}=u_x+v_y$. Hence the isotropic linear-elastic energy density isExpanding the square and collecting terms yields