The gravity wave represents a perturbation around a stationary state, in which there is no velocity. Thus, the perturbation introduced to the system is described by a velocity field of infinitesimally small amplitude, ${\displaystyle (u'(x,z,t),w'(x,z,t)).}$ Because the fluid is assumed incompressible, this velocity field has the streamfunction representation

${\displaystyle {\textbf {u}}'=(u'(x,z,t),w'(x,z,t))=(\psi _{z},-\psi _{x}),\,}$

where the subscripts indicate partial derivatives. In this derivation it suffices to work in two dimensions ${\displaystyle \left(x,z\right)}$, where gravity points in the negative z-direction. Next, in an initially stationary incompressible fluid, there is no vorticity, and the fluid stays irrotational, hence ${\displaystyle \nabla \times {\textbf {u}}'=0.\,}$ In the streamfunction representation, ${\displaystyle \nabla ^{2}\psi =0.\,}$ Next, because of the translational invariance of the system in the x-direction, it is possible to make the ansatz

${\displaystyle \psi \left(x,z,t\right)=e^{ik\left(x-ct\right)}\Psi \left(z\right),\,}$

where k is a spatial wavenumber. Thus, the problem reduces to solving the equation

${\displaystyle \left(D^{2}-k^{2}\right)\Psi =0,\,\,\,\ D={\frac {d}{dz}}.}$

We work in a sea of infinite depth, so the boundary condition is at ${\displaystyle \scriptstyle z=-\infty .}$ The undisturbed surface is at ${\displaystyle \scriptstyle z=0}$, and the disturbed or wavy surface is at ${\displaystyle \scriptstyle z=\eta ,}$ where ${\displaystyle \scriptstyle \eta }$ is small in magnitude. If no fluid is to leak out of the bottom, we must have the condition

${\displaystyle u=D\Psi =0,\,\,{\text{on}}\,z=-\infty .}$

Hence, ${\displaystyle \scriptstyle \Psi =Ae^{kz}}$ on ${\displaystyle \scriptstyle z\in \left(-\infty ,\eta \right)}$, where A and the wave speed c are constants to be determined from conditions at the interface.

The free-surface condition: At the free surface ${\displaystyle \scriptstyle z=\eta \left(x,t\right)\,}$, the kinematic condition holds:

${\displaystyle {\frac {\partial \eta }{\partial t}}+u'{\frac {\partial \eta }{\partial x}}=w'\left(\eta \right).\,}$

Linearizing, this is simply

${\displaystyle {\frac {\partial \eta }{\partial t}}=w'\left(0\right),\,}$

where the velocity ${\displaystyle \scriptstyle w'\left(\eta \right)\,}$ is linearized on to the surface ${\displaystyle \scriptstyle z=0.\,}$ Using the normal-mode and streamfunction representations, this condition is ${\displaystyle \scriptstyle c\eta =\Psi \,}$, the second interfacial condition.

Pressure relation across the interface: For the case with surface tension, the pressure difference over the interface at ${\displaystyle \scriptstyle z=\eta }$ is given by the Young–Laplace equation:

${\displaystyle p\left(z=\eta \right)=-\sigma \kappa ,\,}$

where σ is the surface tension and κ is the curvature of the interface, which in a linear approximation is

${\displaystyle \kappa =\nabla ^{2}\eta =\eta _{xx}.\,}$

Thus,

${\displaystyle p\left(z=\eta \right)=-\sigma \eta _{xx}.\,}$

However, this condition refers to the total pressure (base+perturbed), thus

${\displaystyle \left[P\left(\eta \right)+p'\left(0\right)\right]=-\sigma \eta _{xx}.}$

(As usual, The perturbed quantities can be linearized onto the surface z=0.) Using hydrostatic balance, in the form ${\displaystyle \scriptstyle P=-\rho gz+{\text{Const.}},}$

this becomes

${\displaystyle p=g\eta \rho -\sigma \eta _{xx},\qquad {\text{on }}z=0.\,}$

The perturbed pressures are evaluated in terms of streamfunctions, using the horizontal momentum equation of the linearised Euler equations for the perturbations,

${\displaystyle {\frac {\partial u'}{\partial t}}=-{\frac {1}{\rho }}{\frac {\partial p'}{\partial x}}\,}$

to yield ${\displaystyle \scriptstyle p'=\rho cD\Psi .}$

Putting this last equation and the jump condition together,

${\displaystyle c\rho D\Psi =g\eta \rho -\sigma \eta _{xx}.\,}$

Substituting the second interfacial condition ${\displaystyle \scriptstyle c\eta =\Psi \,}$ and using the normal-mode representation, this relation becomes ${\displaystyle \scriptstyle c^{2}\rho D\Psi =g\Psi \rho +\sigma k^{2}\Psi .}$

Using the solution ${\displaystyle \scriptstyle \Psi =e^{kz}}$, this gives

${\displaystyle c={\sqrt {{\frac {g}{k}}+{\frac {\sigma k}{\rho }}}}.}$