Navier Stokes Equation - Flow of waves

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SUMMARY

The discussion focuses on the Navier-Stokes equation as it pertains to incompressible flow and the derivation of the velocity potential, specifically addressing the equation ##\nabla^2 \phi = 0##. It establishes that for irrotational flow, the velocity can be expressed as the gradient of a potential function, leading to the conclusion that ##\nabla^2 \phi = 0##. Additionally, it outlines boundary conditions for velocity and presents the form of the potential function as ##\phi = -\frac{\omega n_0}{k} cos(kx-\omega t) \frac{cosh kz}{sinh kH}##, with the relationship ##\omega^2 = gk \tan(kH)##. The discussion also raises a question about the frequency variation of waves after a boat is removed from the water.

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This discussion is beneficial for students and professionals in fluid dynamics, particularly those studying the Navier-Stokes equations, as well as researchers focusing on wave behavior in incompressible flows.

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Homework Statement


[/B]
(a) Show that for an incompressible flow the velocity potential satisfies ##\nabla^2 \phi = 0##. Show further the relation for the potential to be ## \frac{\partial \phi}{\partial t} + \frac{\nabla \phi \dot \nabla \phi}{2} + \frac{p}{\rho} + gz = const.##
(b)Write out the boundary conditions on velocity and show ##\phi$## is of the form: ## \phi = -\frac{\omega n_0}{k} cos(kx-\omega t) \frac{cosh kz}{sinh kH}## subject to ## \omega^2 = gk \space tanh \space (kH)##
(c) How does the frequency of the waves vary after the boat is gone?

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Homework Equations

The Attempt at a Solution



Part (a)[/B]

For an irrotational flow, ## \nabla \cdot \vec u = 0## so this means that velocity can be written as gradient of some potential ## u = \nabla \phi##.

Therefore ## \nabla \cdot \vec u = \nabla^2 \phi = 0##.

Navier stokes equation is given by:
\frac{\partial \vec u}{\partial t} + (\vec u \cdot \nabla) \vec u + \frac{1}{\rho} \nabla p + g \vec k = \nu \nabla^2 \vec u

Using the identity and ## \nabla^2 \vec u = 0 ##, ## \nabla \times \vec u \times \vec u = 0## I get the answer.

Part (b)

The boundary condition is velocity at ##z=0## is zero (non-slip).

I don't think this question wants me to simply substitute the given form of ##\phi## to verify it is right. I am expected to derive it.

Substituting, I get
Z \frac{\partial X}{\partial t} + \frac{ (Z\nabla X + X \nabla Z) \cdot (Z\nabla X + X \nabla Z) }{2} + \frac{p}{\rho} + gz = const.

This leads to nowhere. Any idea how to proceed?
 
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bumpp
 
Any help for part (b)??
 
bumpp
 

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