Difference between revisions of "Nonlinear Shallow Water Waves"

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We want to consider waves occurring at the interface of the two fluids water and air. We assume that water is incompressible,
 
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.
 
viscous effects are negligible and that the typical wave lengths are much larger than the water depth.
 +
\vspace{1pt}
  
 
<math>
 
<math>

Revision as of 19:48, 14 October 2008

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. \vspace{1pt}

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

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

Conservation of momentum reads as follows

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

Assuming that changes in the vertical vel. are negligible and [math]\displaystyle{ \vec{u} }[/math] we have [math]\displaystyle{ \frac{D v}{D t} }[/math], thus, [math]\displaystyle{ 0 = \frac{-1}{\rho} }[/math]