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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