= 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 $e_{ij}$, so
$$
\boxed{\Phi=2\mu e_{ij}e_{ij}\ge0}.
$$
For <potential flow>, $u_i=\phi_{,i}$ and mixed <derivatives> commute, giving $e_{ij}=\phi_{,ij}$ and $\Phi=2\mu\sum_{i,j}\phi_{,ij}^2$.
Use the given linear wave potential with $z$ increasing downward and $\theta=kx-\omega t$. The nonzero strain components are
$$
e_{xx}=\omega ak e^{-kz}\sin\theta,\qquad
e_{zz}=-\omega ak e^{-kz}\sin\theta,\qquad
e_{xz}=e_{zx}=\omega ak e^{-kz}\cos\theta.
$$
Their squared contraction is $2\omega^2a^2k^2e^{-2kz}$, independent of the phase. Thus the time-averaged dissipation density is
$$
\boxed{\overline\Phi=4\mu\omega^2a^2k^2e^{-2kz}}.
$$
Depth integration gives dissipation $2\mu\omega^2a^2k$ per unit horizontal area, or $4\pi\mu\omega^2a^2$ per wavelength and unit transverse width.
The velocity components are $u_x=-\omega ae^{-kz}\cos\theta$ and $u_z=\omega ae^{-kz}\sin\theta$. Hence the depth-integrated mean <kinetic energy> per wavelength is
$$
\boxed{\overline K=\frac{\rho}{2}\frac{2\pi}{k}
\int_0^\infty\omega^2a^2e^{-2kz}\,dz
=\frac{\pi\rho\omega^2a^2}{2k^2}}.
$$
Equal mean potential and <kinetic energies> give total <energy> $\mathcal E=\pi\rho\omega^2a^2/k^2$. Dividing the dissipation by that <energy> yields
$$
\boxed{\dot{\mathcal E}=-\frac{4\mu k^2}{\rho}\mathcal E,\qquad
\mathcal E(t)\sim\mathcal E(0)e^{-4\mu k^2t/\rho}}.
$$
The 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.
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