 Solution

For an incompressible Newtonian fluid the viscous stress is ${\tau }_{ij}=2\mu e_{ij}$, where $e_{ij}=(u_{i,j}+u_{j,i})/2$. Its work converted to heat per unit volume is ${\tau }_{ij}{u}_{i,j}$. The antisymmetric part of the velocity gradient has zero contraction with symmetric $e_{ij}$, soFor potential flow, $u_i={\varphi }_{,i}$ and mixed derivatives commute, giving $e_{ij}={\varphi }_{,ij}$ and $\Phi =2\mu {\sum }_{i,j}{\varphi }_{,ij}^2$.

Use the given linear wave potential with $z$ increasing downward and $\theta =kx-\omega t$. The nonzero strain components areTheir squared contraction is $2{\omega }^2{a}^2{k}^2{e}^{-2kz}$, independent of the phase. Thus the time-averaged dissipation density isDepth integration gives dissipation $2\mu {\omega }^2{a}^{2}k$ per unit horizontal area, or $4\pi \mu {\omega }^2{a}^2$ per wavelength and unit transverse width.

The velocity components are $u_x=-\omega a{e}^{-kz}\cos \theta$ and $u_z=\omega a{e}^{-kz}\sin \theta$. Hence the depth-integrated mean kinetic energy per wavelength isEqual mean potential and kinetic energies give total energy $\mathcal{E}=\pi \rho {\omega }^2{a}^2/k^2$. Dividing the dissipation by that energy yieldsThe amplitude decays at half this exponent. The small-viscosity and deep-water assumptions justify evaluating the leading dissipation in the inviscid wave field, extending the depth to infinity and neglecting higher-order amplitude corrections.