Past exam of the options II course of the University of Oxford / 2005 / K / 4

Irrotational motion in this simply connected fluid region has a velocity potential, $u=\nabla \varphi$. Incompressibility gives $\nabla \cdot u=0$, hence $\Delta \varphi =0$.

The moving surface is a material surface: differentiating $y-\eta (x,t)=0$ along a particle gives ${\eta }_t+{\varphi }_x{\eta }_x={\varphi }_y$ at $y=\eta$. Linearizing about a flat stationary surface drops the product term and evaluates at zero, giving ${\eta }_t={\varphi }_y$ there. The Unsteady Bernoulli equation, after dropping quadratic velocity terms and absorbing the time-only constant into $\varphi$, is ${\varphi }_t+p/\rho +gy=0$. The linear capillary pressure at the surface is $p=-T{\eta }_{xx}$ relative to atmosphere, in the sign convention of the source. ThereforeFor $\eta =A\cos (kx-\omega t)$ choose the harmonic potential decaying at infinite depth,Its vertical derivative at zero equals $A\omega \sin (kx-\omega t)={\eta }_t$. Substitution into the dynamic condition gives $-{\omega }^2/k+g+T{k}^2/\rho =0$. ThusThe two positive terms are the gravitational and capillary contributions to the restoring force.