Difference between revisions of "Nonlinear Shallow Water Waves"

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<math>\frac{D \vec{u}}{D t} (\vec{x} ,t) = \nabla p + g(0,-1) </math>
 
<math>\frac{D \vec{u}}{D t} (\vec{x} ,t) = \nabla p + g(0,-1) </math>
  
Assuming that changes in the vertical vel. are negligible and <math> \vec{u} </math> we have <math>\frac{D v}{D t}</math>
+
Assuming that changes in the vertical vel. are negligible and <math> \vec{u} </math> we have <math>\frac{D v}{D t}</math>, thus, <math> 0 = \frac{-1}{\rho} </math>
  
  
 
[[Category:789]]
 
[[Category:789]]

Revision as of 08:05, 14 October 2008

Introduction

We want to consider

[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) = \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]