### Energy dissipation of water waves

This is a interesting, and subtle argument, pointed out to me by Harry Yeh. For a linear wave, energy dissipation rate, averaged over length and time time, is

$2 \mu \frac{1}{LT} \int_0^T\int_0^L\int_{-h}^0{\epsilon_{ij}}\,dydxdt=2 \mu \frac{1}{LT} \int_0^T\int_0^L\int_{-h}^0{\frac{\partial^2 \phi}{\partial x\partial y}^2}\,dydxdt=2 \mu k \frac{a^2 \omega^2}{\tanh k h}$

and it’s apparently non-zero ($\epsilon_{ij}$ is the strain rate tensor). However, for mechanical energy $E(t)$, we have

$E(t)=\int_{D(t)}{\frac{1}{2}\rho |\nabla \phi|^2+\rho g y}\,dV$

and its material derivative is

$\frac{D E}{D t}=\int_D{\rho \nabla \phi\cdot\nabla\phi_t}\,dV+\int_{\partial D}{(\frac{1}{2}\rho|\nabla\phi|^2+\rho g y)U\cdot n}\,ds$

Using $\nabla\cdot(\phi_t\nabla\phi)=\nabla\phi\cdot\nabla\phi_t+\phi_t\nabla^2\phi=\nabla\phi\cdot\nabla\phi_t$ and divergence theorem, we have

$\frac{D E}{D t}=\int_{\partial D}{\rho \phi_t(\nabla\phi\cdot n-U\cdot n)-p U\cdot n}\,ds$

At impermeable bottom, above integral is zero due to $\phi \cdot n=0$ and $U\cdot n=0$; at free surface, kinematic BC and dynamic BC (p=0) indicate above integral is zero as well. Therefore

$DE/Dt=0$

Question: how to explain the seemingly inconsistency of above two results ? i.e. why we have constant mechanical energy, in the meantime a nonzero dissipation rate?

In order to do this, we use the general form of time derivative of mechanical energy $E$:

$\frac{D}{D t}(\rho \frac{|u|^2}{2}+\rho g y)=\nabla\cdot(u\cdot \tau_{ij})-\tau_{ij}:\nabla u$

the 1st term RHS is rate of work done by surroundings to the fluid (material domain), while the 2nd represents the dissipation of mechanical energy. We can see RHS is zero after using divergence theorem and BC’s. But, the key point here is that those two terms are nonzero by themselves, unless flow is inviscid.The last thing to be emphasized is that, the linear momentum for potential flow will have no dissipation term even for viscous flows, due to incompressibility condition.

Now we can see, for a viscous, potential flow, the dissipation rate is non-zero, but its mechanical energy remains unchanged because of the work done at the free surface. This work done, as energy input, is in the form of viscous stress (from air), not covered by DBC $p=0$. If free surface BC is replaced with non-stress condition (different from non-pressure condition), the flow will constantly lose energy due to net dissipation.