Divergence Theorem and Incompressible Fluids

  • Thread starter Bucky
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  • #1
Bucky
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Homework Statement



Hi, I'm trying to follow the proof for the statement
[tex]
\nabla . u = 0
[/tex]

I'm basing it off this paper:
http://delivery.acm.org/10.1145/119...GUIDE&dl=GUIDE&CFID=25582973&CFTOKEN=82107744

(page 7, 8)

In case that's not accessable (I'm in university just now, and I'm not sure if that's a subscriber only paper) I'll write what I've got.


so they start off with defining a fluid volume [tex]\Omega[/tex], and it's boundary surface as [tex]\partial \Omega[/tex], then defining the rate of change around this volume as

[tex] \frac{d}{dt} Volume(\Omega ) = \int \int_{\partial \Omega} u.n [/tex]

The volume should stay constant, thus

[tex] \int \int_{\partial \Omega} u.n = 0 [/tex]

from this step they mention the divergence theorem, then jump to

[tex] \int \int \int_{\Omega} \nabla .u = 0 [/tex]

It's this last jump I don't follow. From http://mathworld.wolfram.com/DivergenceTheorem.html" [Broken], I figured the divergence theorem changed to fit this problem would be..

[tex] \int_{\Omega } (\nabla .u) d\Omega = \int_{\partial \Omega} u.da [/tex]

they've dropped [tex] d\Omega [/tex] and gained two integrals, and I don't follow how they did this.
 
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Answers and Replies

  • #2
Hootenanny
Staff Emeritus
Science Advisor
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I'll admit that their notation isn't really intuitive. Basically, they chose to omit the variables of integration since integration of the surface and volume is implied by the double and triple integrals. However, you should note that what you have,
[tex] \int_{\Omega } (\nabla .u) d\Omega = \int_{\partial \Omega} u.da [/tex]
Is identical to what they have,
[tex] \int \int \int_{\Omega} \nabla .u = 0 [/tex]
Since,
[tex] \int \int_{\partial \Omega} u.n = 0 [/tex]
As I said previously, they have simply chosen not to write [itex]d\Omega[/itex] and [itex]dA[/itex], which is acceptable but can be confusing.
 
Last edited:

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