CFD Online Discussion Forums

CFD Online Discussion Forums (http://www.cfd-online.com/Forums/)
-   Main CFD Forum (http://www.cfd-online.com/Forums/main/)
-   -   what does this mean? \nabla ^3U (http://www.cfd-online.com/Forums/main/124804-what-does-mean-nabla-3u.html)

sharonyue October 13, 2013 21:50

what does this mean? \nabla ^3U
 
\begin{array}{l}
{\nabla ^3}U = ?\\
{\nabla ^2}U = \nabla  \cdot \nabla U
\end{array}

Thanks.

triple_r October 16, 2013 14:37

It should mean:

\nabla^3U=\nabla\bullet\nabla^2U=\nabla\bullet\nabla\bullet\nabla U

Where did you see this term?

sharonyue October 16, 2013 21:20

1 Attachment(s)
Quote:

Originally Posted by triple_r (Post 457323)
It should mean:

\nabla^3U=\nabla\bullet\nabla^2U=\nabla\bullet\nabla\bullet\nabla U

Where did you see this term?

Thanks,

You mean its a scalar rite? Actually its a simple version of the regular one. Its an equation of a turbulence model.

"Partial-average-based equations of incompressible turbulent flow"

I attach a image. This equation is tough to handle. Right now I have no idea how to solve this equation.

Jonas Holdeman October 17, 2013 10:15

It could also be interpreted as a tensor \nabla(\nabla^2 U)

FMDenaro October 17, 2013 12:09

Quote:

Originally Posted by sharonyue (Post 457369)
Thanks,

You mean its a scalar rite? Actually its a simple version of the regular one. Its an equation of a turbulence model.

"Partial-average-based equations of incompressible turbulent flow"

I attach a image. This equation is tough to handle. Right now I have no idea how to solve this equation.


I think is a simbolic power of the operator l.Grad therefore it must be applied n times on the function

triple_r October 18, 2013 11:30

Ignore my first answer please, I didn't see the dot product with l. As Filippo said, it is equal to application of \ell\cdot\nabla n times in a row, and because of the dot product, it is a scalar operator.


All times are GMT -4. The time now is 02:38.