 Solution

The rate-of-strain tensor is $e_{ij}=\frac{1}{2}({\partial }_j{u}_i+{\partial }_i{u}_j)$. For an incompressible Newtonian fluid, the stress tensor is ${\sigma }_{ij}=-p{\delta }_{ij}+2\mu e_{ij}$. Dotting the momentum equation with velocity and using the product rule gives the local mechanical-energy balancewithout body forces. Write ${\partial }_j{u}_i=e_{ij}+{\omega }_{ij}$, with ${\omega }_{ij}$ antisymmetric. The pressure contribution is $-p{\partial }_i{u}_i=0$, and $e_{ij}{\omega }_{ij}=0$ by symmetry. Thus the remaining loss is $2\mu e_{ij}{e}_{ij}\ge 0$, proving the local viscous dissipation rate is $2\mu e_{ij}{e}_{ij}$ per unit volume.

For the oscillating boundary, take $u=(U(y,t),0,0)$ and constant pressure. The nonlinear advection vanishes, leaving $U_t=\nu U_{yy}$, $\nu =\mu /\rho$. Seek the real part of $V{e}^{i\omega t-\gamma y}$; decay at infinity selects $\gamma =(1+i)/\delta$, $\delta =\sqrt{2\nu /\omega }$. After transients,Only $e_{xy}=e_{yx}=U_y/2$ is nonzero, so the dissipation is $\mu U_y^2$. Its time average is $\mu V^2{\delta }^{-2}{e}^{-2y/\delta }$: the two sine and cosine contributions each average to one half. Integrating over $y>0$ gives