Potential theory for water waves

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The discussion focuses on the derivation of the Bernoulli equation for unsteady flow using Newton's second law, specifically addressing the integration process across three dimensions. The confusion arises from how separate Bernoulli equations for each direction are combined into one. It is clarified that since z=0 for the equations in the x and y directions, the results can be merged, as the term involving gravitational potential is constant when z is held constant. The final equation is valid across all three dimensions, illustrating a fundamental concept in potential theory. This integration process is likened to finding a function from its partial derivatives.
Nikitin
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hi :)

http://ocw.mit.edu/courses/mechanic...s-fall-2009/course-text/MIT2_017JF09_ch06.pdf

In page 37 they use Newton's 2nd law for a fluid element (while ignoring viscous forces) to derive the bernoulli equation for unsteady flow.

Well, what I am confused about is the last step: They integrate through the x direction, y direction and then z direction to get 3 separate bernoulli equations,,, but then they just add it all up to one? I don't understand the mathematics behind this as I'm completely new to potential theory.
 
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Furthermore, on page 40 they equate the dynamical pressure {for waterwaves} to be equal to the change in the potential function?? To see what I mean:

From this (Bernoulli) ##p + 0.5 \rho V^2 (\approx 0) + \rho g z +\rho \frac{\partial(\phi)}{\partial(z)} = C##, they go to this: ##p_d = p+\rho g z = -\rho \frac{\partial(\phi)}{\partial(z)}##

How did they do that?
 
Nikitin said:
hi :)

http://ocw.mit.edu/courses/mechanic...s-fall-2009/course-text/MIT2_017JF09_ch06.pdf

In page 37 they use Newton's 2nd law for a fluid element (while ignoring viscous forces) to derive the bernoulli equation for unsteady flow.

Well, what I am confused about is the last step: They integrate through the x direction, y direction and then z direction to get 3 separate bernoulli equations,,, but then they just add it all up to one? I don't understand the mathematics behind this as I'm completely new to potential theory.

This is nothing mysterious, it is just that z=0 for the equations in \hat{x} and \hat{y}, so that equation works for all three cases. If that doesn't make sense, do the three integrals and compare the results.
 
It works through "The first force balance (x direction)" and gets to the result

"Bermouilli forrmula" = ##C_1## when y and z are constant

The same process in the y direction gives
"Bermouilli forrmula" = ##C_2## when z and x are constant

and in the z direction
"Bermouilli forrmula" + ##\rho g z## = ##C_3## when x and y are constant

But ##\rho g z## is a constant when z is constant, so you can merge it into ##C_1## and ##C_2##.

In other words the final equation is actually valid in the x, y, and z directions.

This seems like a tortuous way to get to the result, but I guess that depends what you already know and what the teaching objectives for this part of the course are.
 
Oh how stupid of me. This is just the same as integrating partial derivatives of a function and then comparing the results to find the actual function. OK, thanks!
 
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