Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 4 / 38D (Fluid Dynamics)

The momentum equation isTaking its scalar product with $u$ givesFor the Newtonian fluid stress tensor $\sigma =-pI+2\mu e$, its contraction with the velocity gradient isbecause incompressibility removes the pressure term and the symmetric rate-of-strain tensor is orthogonal to the antisymmetric part of $\nabla u$. Integration over $\mathcal{D}$ and the divergence theorem therefore give the kinetic-energy balance for an incompressible Newtonian fluid:The terms are respectively the rate of change of kinetic energy, outward advective flux of kinetic energy, power supplied by the body force, power supplied by surface traction, and irreversible viscous dissipation. Since $f=-\nabla \psi$, the body-force power may also be written $-\nabla \cdot (\psi u)$ and interpreted as exchange with potential energy.

The linearized kinematic condition at the undisturbed surface isFor the given complex amplitudes this says

The velocity is $u=({\varphi }_x,0,{\varphi }_z)$. Its nonzero complex strain amplitudes areUsing the stated rule for period averages,Thus the mean dissipation per unit horizontal area isThe mean energy $E=\rho g|A{|}^2/2$ consequently obeys $dE/dt=-\mathcal{D}$. Writing $\nu =\mu /\rho$ gives the viscous decay of a deep-water gravity wave: