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Take the outward wavenumber $k>0$; the absolute value in the source makes the two propagation directions equivalent. The phase velocity of the gravity-capillary wave obeysDifferentiate with respect to $k$:The second derivative is positive, and $c_p\to \infty$ at both ends of the positive-$k$ range, so this is the global minimum. The minimum phase speed of a gravity-capillary wave isFor $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ and the supplied surface tension and mass density,The corresponding wavenumber is $k_{\ast }\simeq 369{\mathrm{m}}^{-1}$, giving wavelength ${\lambda }_{\ast }=2\pi /k_{\ast }\simeq 1.70\mathrm{cm}$. At this minimum, $c_g=d\omega /dk=c_p+kd{c}_p/dk=c_p$. Thus the group velocity and phase velocity agree for this particular ripple, although they differ for general gravity-capillary waves.

Put $\tau =T/\rho >0$ and take the positive-frequency branch for $k>0$. The phase velocity and group velocity areSince the square root is increasing, minimizing $c_p$ is equivalent to minimizing $g/k+\tau k$. Its derivative is $-g/k^2+\tau$, which changes sign from negative to positive atThe phase speed diverges at both positive endpoints $k\downarrow 0$ and $k\to \infty$, so this stationary point is its global minimum. At $k_c$, $\tau k_c^2=g$, givingThe relative speed is especially transparent:A narrow wave packet moves with $c_g$, while its crests move with $c_p$. For $k<k_c$, $c_p>c_g$: crests move forward through the packet, from its rear towards its front. For $k>k_c$, $c_p<c_g$: crests drift backwards relative to the packet, although both absolute velocities are positive. At $k_c$ the leading-order relative drift vanishes. This is the minimum phase speed of a gravity-capillary wave.

The free surface is material, so applying the material derivative to $y-\eta (x,t)$ gives the exact kinematic boundary conditionFor an incompressible fluid with constant density, the Euler equations areSince the flow is irrotational, $\mathbf{u}=\nabla \varphi$ and $(\mathbf{u}\cdot \nabla )\mathbf{u}=\nabla (|\nabla \varphi {|}^2/2)$. Integrating the resulting gradient equation gives a function of time on the right of Bernoulli equation. Adding a time-dependent constant to $\varphi$ removes that function, without changing velocity, and choosing the atmospheric-pressure reference givesThe 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 ${\varphi }_x{\eta }_x$ and $|\nabla \varphi {|}^2$, the slope-squared correction to curvature, and displacement corrections in evaluating $\varphi$ at $y=\eta$ are all higher order. Hence, on $y=0$,For $\eta =A\cos (kx-\omega t)$, a harmonic velocity potential decaying as $y\to -\infty$ and satisfying the first condition isIndeed $\Delta \varphi =0$ and ${\varphi }_y{|}_0=A\omega \sin (kx-\omega t)={\eta }_t$. Substituting in the second condition yields the dispersion relation for a gravity-capillary wave:Taking the positive-frequency branch, the phase velocity and group velocity areSubtracting givesSince $\omega >0$,The threshold is also the wavenumber at the minimum phase speed of a gravity-capillary wave.

For a monochromatic wave, a constant phase $kx-\omega t$ moves at phase velocity $v_p=\omega /k$. For a narrow wave packet around $k$, expansion $\omega (k+\Delta k)\simeq \omega (k)+(d\omega /dk)\Delta k$ shows that its envelope moves at group velocity $v_g=d\omega /dk$. The envelope interpretation requires a sufficiently narrow spectrum and weak distortion; it is not a general identification with a causal signal front. Since $\omega =k{v}_p$ and $\lambda =2\pi /k$,For the gravity-capillary wave, put $s=\sigma /\rho$ and take $k>0$. Direct differentiation givesThus $v_g=v_p$ when $s{k}_{\ast }^2=g$, orAt this wavelength the gravity and surface-tension restoring terms contribute equally. Also $d(v_p^2)/dk=-g/k^2+s=0$ with positive second derivative: this is the minimum phase speed of a gravity-capillary wave, $v_{p,\min }=(4g\sigma /\rho {)}^{1/4}=0.214\mathrm{m}{\mathrm{s}}^{-1}$. Locally the carrier crests and narrow envelope travel together. The medium has not become nondispersive: $d^2\omega /d{k}^2=k_{\ast }{d}^2{v}_p/d{k}^2\ne 0$ there, so a finite wave packet can still spread. In the gravity-dominated limit $v_g/v_p\to 1/2$; in the capillary-dominated limit it tends to $3/2$.