Difference between revisions of "Nonlinear Shallow Water Waves"

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</math>
 
</math>
  
Since water is incompressible i.e. <math>  \frac{D \rho}{D t} = 0  </math> and then <math>\nabla \cdot u = 0</math>
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Since water is incompressible i.e. <math>  \frac{D \rho}{D t} = 0  </math> and then <math>\nabla \cdot \vec{u} = 0</math>
  
 
[[Category:789]]
 
[[Category:789]]

Revision as of 07:42, 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]