- #1
Bucky
- 82
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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" , 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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