# Question about Airy wave theory

I have a fundamental misunderstanding about Airy wave theory. All the texts that derive the dispersion relation for gravity water waves use as a final step the unsteady Bernoulli equation: $$\frac{\mathrm d\Phi}{\mathrm dt} + g\eta = 0 .$$ But as far as I know, the unsteady Bernoulli equation also includes a squared-velocity term (and a pressure term that is constant), so it must indeed by written as $$\frac{\mathrm d\Phi}{\mathrm dt} + \frac12(v_x^2 + v_y^2) + g\eta = 0.$$ Now comes the point: if the fluid particles move in circles in constant speed, then the term $$\frac12(v_x^2 + v_y^2)$$ is also constant (like the atmospheric pressure) so the Bernoulli equation reduces to the familiar equation.

This assumption is indeed correct in deep water, but in intermediary water depth and shallow water it's not correct, but I have seen cases in which the Airy wave theory is also applied in this regimes. So what did I miss?

• The reason we ignore this term is because it is smaller than the linear terms, where the small parameter is the wave slope $ak$. I am not sure what point you're making wrt to the kinetic energy term (you need to do the asymptotics formally to get a rational solution). See my answer here for more details: physics.stackexchange.com/questions/1289/… – Nick P Aug 25 '17 at 2:20

O.K i think Nick P has explained to me the point very clearly. The velocity term in the unsteady Bernoulli equation is neglected for two reasons: the first and more important is that for waves with small slope $$ak$$ the velocity of the fluid elements is small in comparison with the wave speed (the ratio of their speed is equal to the wave slope), and since the wave speed is also not that high then the fluid elements speed is neglible , so the only important terms in Bernoulli equation are only the time derivative of the velocity potential and $$g\eta$$. The assumption that wave slope is small is the central linearization assumption of Airy wave theory, and is nessecery in order for that the Stokes drift will not be two large.