Nonlinear Shallow Water Waves

Introduction

We want to consider waves occurring at the interface of the two fluids water and air. We assume that water is incompressible, viscous effects are negligible and that the typical wave lengths are much larger than the water depth. For "shallow" water we assume no variations in the y-dirn. Let the domain of interest be denoted by [math]\displaystyle{ \Omega }[/math] with z-axis vertical and x-axis horizontal, [math]\displaystyle{ \vec{x} = (x,z) }[/math]. Let [math]\displaystyle{ h_{o} }[/math] be the at rest water depth. Let [math]\displaystyle{ h(x,t) }[/math] be the local water depth or wave height. From previous lectures we have established the eqn. for Conservation of Mass.

[math]\displaystyle{ \frac{D \rho}{D t} (\vec{x} ,t) + \rho(\vec{x} ,t)\nabla \cdot \vec{u}(\vec{x} ,t) = 0, x \in \Omega }[/math]

Since water is incompressible [math]\displaystyle{ \frac{D \rho}{D t} = 0 }[/math] i.e. [math]\displaystyle{ \nabla \cdot \vec{u} = 0 }[/math], the divergance of the velocity field is zero.

We can apply the Conservation of Momentum eqn. as follows

[math]\displaystyle{ \frac{D \vec{u}}{D t} (\vec{x} ,t) = \frac{-1}{\rho} \nabla p + g(0,-1) }[/math]

components of [math]\displaystyle{ \vec{u} = (u,v) }[/math], assuming that changes in the vertical velocity are negligible i.e,[math]\displaystyle{ \frac{D v}{D t} (\vec{x} ,t) = 0 }[/math]

thus, [math]\displaystyle{ 0 = \frac{-1}{\rho}\frac{\partial p}{\partial z} - g }[/math] or [math]\displaystyle{ \frac{\partial p}{\partial z} = -\rho g }[/math] and hence,

[math]\displaystyle{ p(\vec{x} ,t) = p_{atm} + \rho g(h(x,t) - z) }[/math] (pressure is hydrostatic)

Also, [math]\displaystyle{ \frac{D u}{D t}(\vec{x} ,t) = \frac{-1}{\rho}\frac{\partial p}{\partial x} }[/math], and since [math]\displaystyle{ \frac{\partial p}{\partial x} }[/math] is independant of [math]\displaystyle{ z }[/math], so is [math]\displaystyle{ \frac{D u}{D t} }[/math]

[math]\displaystyle{ \frac{D u}{D t} = -g\frac{\partial h(x ,t)}{\partial x} }[/math] after substituting expression for [math]\displaystyle{ p(\vec{x} ,t) }[/math]

Therefore if [math]\displaystyle{ u }[/math] has no z-dependance at the, at rest condition ([math]\displaystyle{ h_{o} }[/math]), it will remain this way. Thus,

[math]\displaystyle{ \frac{\partial u(\vec{x} ,t)}{\partial t} + u(\vec{x} ,t)\frac{\partial u(\vec{x} ,x)}{\partial x} + g \frac{\partial h(x ,t)}{\partial x} = 0 }[/math]

Another equation is needed. Consider conservation of mass along a vertical cross section.

[math]\displaystyle{ \frac{d}{dt}\int_{x_1}^{x_2} \rho h(x ,t) dx = -[\rho u(\vec{x} ,t) h(x ,t)]_{x_1}^{x_2} }[/math]

Assuming quantities are smooth and [math]\displaystyle{ x_1, x_2 }[/math] are arbitrary we arrive at

[math]\displaystyle{ \frac{\partial h(x ,t)}{\partial t} + \frac{\partial u(\vec{x} ,x)h(x ,t)}{\partial x} = 0 }[/math]