Past exam of the options II course of the University of Oxford / 2004 / L / L2 / Solution

The velocity is $\mathbf{u}=\nabla \varphi$ because the flow is irrotational. Incompressible flow requires $\nabla \cdot \mathbf{u}=0$, hence the velocity potential satisfies Laplace's equation:The exact kinematic boundary condition at the free surface is ${\eta }_t+{\varphi }_x{\eta }_x={\varphi }_y$. With constant atmospheric pressure, the dynamic condition from Bernoulli's principle is ${\varphi }_t+\frac{1}{2}|\nabla \varphi {|}^2+g\eta =0$, after choosing the time-dependent potential gauge. For small amplitudes, the products of perturbations are second order. Evaluating the remaining first-order terms at the undisturbed level $y=0$ givesSubstitution of $\varphi =f(y)\sin (kx-\omega t)$ into Laplace's equation gives $f^{\prime \prime }-k^{2}f=0$. For $k\ne 0$, decay as $y\to -\infty$ selects $f(y)=B{e}^{|k|y}$. The two boundary conditions for $\eta =A\cos (kx-\omega t)$ yieldFor a nonzero wave amplitude these are compatible precisely whenWith the conventional $k>0$ and positive frequency, the dispersion relation is $\omega (k)=\sqrt{gk}$. The exponential decay is the infinite-depth boundary condition; retaining a growing exponential would violate it.