Depth-dependent ocean surface wave solution

Susan Hautala

69 Depth-dependent ocean surface wave solution

We will assume a wave propagating in the x direction, with no variation in the y direction. This is sometimes called a plane wave, since motion occurs in a 2D plane.

We assume no friction, which implies no vorticity added to the fluid from solid boundaries,

$\nabla \times \textbf{\underline{u}}=\frac{\partial w}{\partial x}-\frac{\partial u}{\partial z}=0.$

For a system with no vorticity, the velocity field can be expressed as the gradient of a scalar, $\Phi$ , called the velocity potential,

$u = \frac{\partial \Phi}{\partial x} \quad w = \frac{\partial \Phi}{\partial z}.$

(You can prove to yourself that this is true by expressing the vorticty in terms of $\Phi.$ ) The simplest way to find the ocean surface wave solution is to use this velocity potential. We will assume a wave that is propagating in the horizontal direction, but with a depth dependent velocity potential amplitude,

$\Phi = a(z) \hspace{2pt} cos(kx-\omega t).$

The continuity equation ( $\nabla \cdot \textbf{\underline{u}}=0$ ) can be written in terms of the velocity potential as

$\frac{\partial^2 \Phi}{\partial x^2}+\frac{\partial^2 \Phi}{\partial z^2}=-k^2a+\frac{\partial^2 a}{\partial z^2}=0$

Notice the difference in sign relating the function, $a(z)$ to its second derivative, compared to the equation for a harmonic oscillator. It is easy to show that a solution that also satisfies the boundary condition $w= \partial \Phi / \partial z = 0$ at the seafloor z = -h is a hyperbolic cosine, $a(z)=Acosh[k(z+h)].$

The boundary condition at the free surface is a little bit more complicated. Two equations have to be satisfied:

(1) The height of the sea surface is a function of time with its rate of change equal to the vertical velocity of water parcels,

$\frac{\partial \eta}{\partial t}=w=\frac{\partial \Phi}{\partial z}$

(2) The vertical momentum equation relating vertical acceleration to combined pressure gradient and gravitational forces (we have relaxed the hydrostatic pressure assumption, but we still assume small amplitude waves so that the inertial term is small)

$\frac{\partial w}{\partial t}=\frac{\partial^2 \Phi}{\partial t \partial z}=-\frac{1}{\rho_0}\frac{\partial P}{\partial z}-g,$

which can be integrated with respect to z,

$\frac{\partial \Phi}{\partial t}=-\frac{P}{\rho_0}-gz=-g \eta.$

where we have used the values at the sea surface where $z=\eta$ and $P=0$ .

We combine these two equations to get

$\frac{\partial^2 \Phi}{\partial t^2}=-g \frac{\partial \Phi}{\partial z}.$

Substituting in our function for $\Phi=Acosh[k(z+h)]cos(kx-\omega t)$ , and cancelling out terms that appear on both sides, this equation becomes, at the surface, z = 0 (again assuming a small amplitude wave),