= Solution
The <free surface> is material, so applying the material <derivative> to $y-\eta(x,t)$ gives the exact <kinematic boundary condition>
$$
\boxed{\eta_t+\phi_x\eta_x=\phi_y\quad\text{at }y=\eta.}
$$
For an <incompressible flow>[incompressible fluid] with constant density, the <Euler equations> are
$$
\partial_t\boldsymbol u+(\boldsymbol u\cdot\nabla)\boldsymbol u
=-\rho^{-1}\nabla p-g\boldsymbol e_y.
$$
Since the flow is <irrotational flow>[irrotational], $\boldsymbol u=\nabla\phi$ and $(\boldsymbol u\cdot\nabla)\boldsymbol u=\nabla(|\nabla\phi|^2/2)$. Integrating the resulting <gradient> equation gives a function of time on the right of <Bernoulli equation>. Adding a time-dependent constant to $\phi$ removes that function, without changing velocity, and choosing the atmospheric-pressure reference gives
$$
\phi_t+\frac12|\nabla\phi|^2+gy+\frac{p-p_a}\rho=0.
$$
The pressure immediately above the surface is atmospheric; the liquid-side pressure includes the stated <surface tension> jump.
At first order in wave amplitude, the products $\phi_x\eta_x$ and $|\nabla\phi|^2$, the slope-squared correction to curvature, and displacement corrections in evaluating $\phi$ at $y=\eta$ are all higher order. Hence, on $y=0$,
$$
\phi_y=\eta_t,\qquad
\phi_t+g\eta-\frac T\rho\eta_{xx}=0.
$$
For $\eta=A\cos(kx-\omega t)$, a harmonic <velocity potential> decaying as $y\to-\infty$ and satisfying the first condition is
$$
\phi=\frac{A\omega}{k}e^{ky}\sin(kx-\omega t).
$$
Indeed $\Delta\phi=0$ and $\phi_y|_0=A\omega\sin(kx-\omega t)=\eta_t$. Substituting in the second condition yields the <dispersion relation> for a <gravity-capillary wave>:
$$
\boxed{\omega^2=gk+\frac T\rho k^3.}
$$
Taking the positive-frequency branch, the <phase velocity> and <group velocity> are
$$
c=\frac\omega k=\sqrt{\frac gk+\frac T\rho k},
\qquad
c_g=\frac{d\omega}{dk}=\frac{g+3Tk^2/\rho}{2\omega}.
$$
Subtracting gives
$$
c_g-c=\frac{Tk^2/\rho-g}{2\omega}.
$$
Since $\omega>0$,
$$
\boxed{c_g>c\iff k>\sqrt{\rho g/T}.}
$$
The threshold is also the wavenumber at the <minimum phase speed of a gravity-capillary wave>.
\Image[/past-exam-of-the-options-ii-course-of-the-university-of-oxford/2011-capillary-gravity-speeds.png]
{title=Phase and group speeds of deep-water gravity-capillary waves}
{description=The curves meet at dimensionless wavenumber one, where the phase speed is minimal.}
Back to article page
