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A gravity-capillary wave has restoring contributions from both gravity and surface tension. Consider an initially resting, horizontally unbounded liquid layer of constant depth $h$, with uniform mass density $\rho$, incompressible flow, inviscid flow and irrotational flow. The upper gas exerts constant pressure and has negligible inertia. Assume a clean interface with constant surface tension $\gamma$: no surfactant stresses, imposed current or wind. The wave amplitude $a$ must satisfy $ka\ll 1$ and $a\ll h$, so linearized boundary conditions apply. Deep water means $kh\gg 1$, rather than a particular depth in metres. ^[ref]

Let $z=0$ be the undisturbed free surface, with liquid below it. Potential flow gives a velocity potential $\varphi$ with $\mathbf{u}=\nabla \varphi$, and incompressible flow gives the Laplace equation. The linearized kinematic boundary condition for a free-surface graph and the Unsteady Bernoulli equation, including gravity and the Young–Laplace equation, giveHere $\eta (x,t)$ is the interface displacement; the signed small-slope pressure jump is $p_{\mathrm{liquid}}-p_{\mathrm{gas}}=-\gamma {\eta }_{xx}$. A wave crest has ${\eta }_{xx}<0$, so its capillary pressure is positive. Choose a right-travelling Fourier mode of positive wavenumber $k$ and angular frequency $\omega$:The factor $e^{kz}$ decays into the deep liquid, satisfies the Laplace equation, and makes ${\varphi }_z={\eta }_t$ at $z=0$. Substituting in the dynamic boundary condition leavesThis derivation displays the two restoring contributions with the same sign. Dimensional analysis also checks the result: $\gamma /\rho$ has units ${\mathrm{m}}^3{\mathrm{s}}^{-2}$, so both $gk$ and $(\gamma /\rho )k^3$ have units ${\mathrm{s}}^{-2}$. ^[ref]

At finite depth, a flat impermeable bottom at $z=-h$ requires ${\varphi }_z=0$. Replacing $e^{kz}$ by the normalized vertical factor $\cosh (k(z+h))/\cosh (kh)$ makes the surface ratio ${\varphi }_z/\varphi =k\tanh (kh)$. Eliminating the mode amplitudes then givesThus the same model supplies its own depth check: use the deep-water formula only when $\tanh (kh)$ is sufficiently close to one for the required accuracy. For example, $kh=3$ reduces $\omega$ by about $0.248\%$ relative to deep water, while $kh=1$ reduces it by about $12.7\%$. These percentages are calculations from this factor. ^[ref]

In deep water the phase velocity magnitude follows a wave crest, while the group velocity magnitude follows the wave envelope of a narrow wave packet. Differentiating the positive-frequency dispersion relation givesFor $k{\ell }_c\ll 1$, the surface gravity wave limit has $c_g\simeq c_p/2$. For $k{\ell }_c\gg 1$, the deep-water capillary-wave dispersion relation has $c_g\simeq 3c_p/2$: a wave crest slips backward relative to the faster wave envelope, even while both advance. Neither speed is the oscillating fluid element's own velocity. ^[ref]

There is a particularly neat coincidence at the crossover. Differentiate $c_p^2=g/k+\gamma k/\rho$: its unique minimum is at $k_c=\sqrt{\rho g/\gamma }=1/{\ell }_c$, exactly where the restoring terms are equal. ThereforeFor the stipulated parameters in capillary length, the minimum is $0.231\mathrm{m}{\mathrm{s}}^{-1}$. At $\lambda =3\mathrm{mm}$, direct substitution gives $f=\omega /(2\pi )\simeq 131\mathrm{Hz}$, $c_p\simeq 0.394\mathrm{m}{\mathrm{s}}^{-1}$ and $c_g\simeq 0.580\mathrm{m}{\mathrm{s}}^{-1}$. This short wavelength is capillary dominated, but its group velocity is still slightly below the pure-capillary ratio $3c_p/2$ because gravity has not vanished.

Linearize about the resting state $\eta =0$, $\varphi =0$. Introduce a dimensionless small parameter $ϵ$ and writeholding the depth, bending stiffness and variation scales fixed. Thus both boundary displacement and induced velocity are small. For any first-order derivative $D\varphi$, a Taylor expansion at the moving surface givesThe displacement correction to the leading field is ${ϵ}^2{\eta }_1{\partial }_z(D{\varphi }_1)(x,0,t)$, so evaluating it on $z=0$ changes it only at second order. Also ${\varphi }_x{\eta }_x=O({ϵ}^2)$ and $|\nabla \varphi {|}^2=O({ϵ}^2)$. The first-order kinematic boundary condition and dynamic boundary condition are thereforeDropping the subscript, the linearized boundary condition pair isThe fixed-bed condition remains ${\varphi }_z(x,-h,t)=0$. The source's shorthand $|\eta |\ll 1$ requires a length scale: for the subsequent plane wave, small surface slope $k|\alpha |\ll 1$ and displacement small compared with depth are sufficient conditions. Small height alone, with arbitrarily rapid spatial variation or a finite background disturbance, does not justify discarding the products.

Use the linearized boundary condition: evaluate at $z=0$ and discard products of perturbations. The kinematic boundary condition and Young–Laplace equation becomeIncompressible flow and irrotational flow give the Laplace equation for each potential. For a normal mode, the wall conditions selectWith $\eta =B{e}^{i(kx-\omega t)}$, the kinematic conditions give $A_1=i\omega B/[k\sinh (k{L}_1)]$ and $A_2=-i\omega B/[k\sinh (k{L}_2)]$. Hence ${\hat{\varphi }}_1(0)=i\omega B\coth (k{L}_1)/k$ and ${\hat{\varphi }}_2(0)=-i\omega B\coth (k{L}_2)/k$. Substitution into the dynamic condition gives the finite-depth Rayleigh-Taylor dispersion relation:Its denominator is positive. Assuming $g>0$ and $\gamma >0$, negative ${\omega }^2$ therefore produces a growing branch $\omega =i\sigma$ with $\sigma >0$, as in Rayleigh-Taylor instability.