what does this mean? \nabla ^3U

sharonyue · October 13, 2013, 21:50

$\begin{array}{l} {\nabla ^3}U = ?\\ {\nabla ^2}U = \nabla \cdot \nabla U \end{array}$

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

Quote:

Originally Posted by triple_r

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

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.

October 13, 2013, 21:50	what does this mean? \nabla ^3U	#1
sharonyue Senior Member Dongyue Li Join Date: Jun 2012 Location: Beijing, China Posts: 848 Rep Power: 18	$\begin{array}{l} {\nabla ^3}U = ?\\ {\nabla ^2}U = \nabla \cdot \nabla U \end{array}$ Thanks.

October 16, 2013, 14:37		#2
triple_r Senior Member Reza Join Date: Mar 2009 Location: Appleton, WI Posts: 116 Rep Power: 17	It should mean: $\nabla^3U=\nabla\bullet\nabla^2U=\nabla\bullet\nabla\bullet\nabla U$ Where did you see this term? Last edited by triple_r; October 16, 2013 at 14:37. Reason: formatting the equation correctly

October 18, 2013, 11:30		#6
triple_r Senior Member Reza Join Date: Mar 2009 Location: Appleton, WI Posts: 116 Rep Power: 17	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. FMDenaro likes this.

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Jonas Holdeman Senior Member Jonas T. Holdeman, Jr. Join Date: Mar 2009 Location: Knoxville, Tennessee Posts: 128 Rep Power: 18	It could also be interpreted as a tensor $\nabla(\nabla^2 U)$