Difference between revisions of "Nonlinear Shallow Water Waves"
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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> i.e. the divergance of the velocity field is zero. | Since water is incompressible i.e. <math> \frac{D \rho}{D t} = 0 </math> and then <math>\nabla \cdot \vec{u} = 0</math> i.e. the divergance of the velocity field is zero. | ||
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| + | Conservation of momentum reads as follows | ||
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[[Category:789]] | [[Category:789]] | ||
Revision as of 07:46, 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
