| Consider two fluid domains, separated by an interface with surface tension. The mean interface position is horizontal. It separates the upper from the lower fluid, both having a different constant mass density, ρ and ρ’ for the lower and upper domain respectively. The fluid is assumed to be inviscid and incompressible, and the flow is assumed to be irrotational. Then the flows are potential, and the velocity in the lower and upper layer can be obtained from ∇Φ and ∇Φ’, respectively. Here Φ(x,y,z,t) and Φ’(x,y,z,t) are velocity potentials.
Three contributions to the energy are involved: the potential energy Vg due to gravity, the potential energy Vst due to the surface tension and the kinetic energy T of the flow. The part Vg due to gravity is the simplest: integrating the potential energy density due to gravity, ρ g z (or ρ’ g z) from a reference height to the position of the surface, z = η(x,y,t)[5]:

assuming the mean interface position is at z=0.
An increase in area of the surface causes a proportional increase of energy due to surface tension[6]:
![V_\mathrm{st}
= \sigma \iint dx\, dy\;
\left[
\sqrt{ 1 + \left( \frac{\partial \eta}{\partial x} \right)^2
+ \left( \frac{\partial \eta}{\partial y} \right)^2}
- 1
\right]
\approx \frac{1}{2} \sigma \iint dx\, dy\;
\left[
\left( \frac{\partial \eta}{\partial x} \right)^2
+
\left( \frac{\partial \eta}{\partial y} \right)^2
\right],](http://wpcontent.answers.com/math/d/6/c/d6c7638383a793b705c0e7d820246bf0.png)
where the first equality is the area in this (Monge's) representation, and the second applies for small values of the derivatives (surfaces not too rough).
The last contribution involves the kinetic energy of the fluid[7]:
![T=
\frac{1}{2} \iint dx\, dy\;
\left[
\int_{-\infty}^\eta dz\; \rho\, \left| \bold\nabla \Phi \right|^2
+
\int_\eta^{+\infty} dz\; \rho'\, \left| \bold\nabla \Phi' \right|^2
\right].](http://wpcontent.answers.com/math/5/1/c/51ce2ace5a7f9371a4c053301af8339e.png)
Use is made of the fluid being incompressible and its flow is irrotational (often, sensible approximations). As a result, both Φ(x,y,z,t) and Φ’(x,y,z,t) must satisfy the Laplace equation[8]:
and 
These equations can be solved with the proper boundary conditions: Φ and Φ’ must vanish well away from the surface (in the "deep water" case, which is the one we consider).
Using Green's identity, and assuming the deviations of the surface elevation to be small (so the z–integrations may be approximated by integrating up to z=0 instead of z=η), the kinetic energy can be written as[7]:

To find the dispersion relation, it is sufficient to consider a sinusoidal wave on the interface, propagating in the x–direction[6]:

with amplitude a and wave phase θ = kx - ωt. The kinematic boundary condition at the interface, relating the potentials to the interface motion, is that the vertical velocity components must match the motion of the surface[6]:
and at z = 0.
To tackle the problem of finding the potentials, one may try separation of variables, when both fields can be expressed as[6]:

Then the contributions to the wave energy, horizontally integrated over one wavelength λ = 2π/k in the x–direction, and over a unit width in the y–direction, become[6][9]:

The dispersion relation can now be obtained from the Lagrangian L = T - V, with V the sum of the potential energies by gravity Vg and surface tension Vst[10]:
![L = \frac{1}{4} \left[
(\rho+\rho') \frac{\omega^2}{|k|} - (\rho-\rho') g - \sigma k^2
\right] a^2 \lambda.](http://wpcontent.answers.com/math/f/4/1/f415d03b8d891e98718a18a332945126.png)
For sinusoidal waves and linear wave theory, the phase–averaged Lagrangian is always of the form L = D(ω,k) a², so that variation with respect to the only free parameter, a, gives the dispersion relation D(ω,k) = 0 [10]. In our case D(ω,k) is just the expression in the square brackets, so that the dispersion relation is:

the same as above.
As a result, the average wave energy per unit horizontal area, ( T + V ) / λ, is:
![\bar{E} = \frac{1}{2}\, \left[ (\rho-\rho')\, g + \sigma k^2 \right]\, a^2.](http://wpcontent.answers.com/math/a/8/8/a88a36ca97c60877e3b1ac54a19403bc.png)
As usual for linear wave motions, the potential and kinetic energy are equal (equipartition holds): T = V.[11]
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