Difference between revisions of "Nonlinear Shallow Water Waves"

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<math>(M_{b} + M_{s}) \frac{d^2 x_s}{d t^2} + M_{b} L_{s} cos{\theta_s} \frac{d^2 \theta_s}{d t^2} + M_{b} L_{b} cos{\theta_b} \frac{d^2 \theta_b}{d t^2} - M_{b} L_{s} sin{\theta_s} (\frac{d \theta_s}{d t})^2</math>
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<math>(M_{b} + M_{s}) \frac{d^2 x_s}{d t^2} + M_{b} L_{s} cos{\theta_s} \frac{d^2 \theta_s}{d t^2} + M_{b} L_{b} cos{\theta_b} \frac{d^2 \theta_b}{d t^2} - M_{b} L_{s} sin{\theta_s} (\frac{d \theta_s}{d t})^2 - M_{b} L_{b} sin{\theta_b} (\frac{d \theta_b}{d t})^2</math>
  
  
  
 
[[Category:789]]
 
[[Category:789]]

Revision as of 22:44, 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. 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]


Assuming that changes in the vertical velocity are negligible


and [math]\displaystyle{ \vec{u} = (u,v) }[/math], we have [math]\displaystyle{ \frac{D v}{D t} }[/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 = p_{atm} + \rho g(h(x,t) - z) }[/math] (pressure is hydrostatic)


Also, [math]\displaystyle{ \frac{D u}{D 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]

Therefore if [math]\displaystyle{ u }[/math] has no


[math]\displaystyle{ (M_{b} + M_{s}) \frac{d^2 x_s}{d t^2} + M_{b} L_{s} cos{\theta_s} \frac{d^2 \theta_s}{d t^2} + M_{b} L_{b} cos{\theta_b} \frac{d^2 \theta_b}{d t^2} - M_{b} L_{s} sin{\theta_s} (\frac{d \theta_s}{d t})^2 - M_{b} L_{b} sin{\theta_b} (\frac{d \theta_b}{d t})^2 }[/math]