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Question about Divergence Operation (del . u==u . del ?)

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Hi, I have a question about the del operator (which is upside down delta). It's a vector containing incomplete partial differential, isn't it ? del = (d/dx, d/dy, d/dz) As for the divergece, it's a dot product... suppose we have v = (vx, vy, vz), then del . v = dvx/dx + dvy/dy + dvz/vz My question, is, is del . v = v . del ? I wonder, since I found navier stokes equation : du/dt = -(u . del)u - v del^2 u + del p + f And, to make it divergence free, we have the second equation : del . u = 0 For now, I think that del . u is same as u . del, doesn't that make the navier-stokes equation become : du/dt = - v del^2 + del p + f ??? Thanks.

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No, they aren't...
If I assume that u = (ux, uy, uz)
then
[ du/dt = -(u . del)u - v del^2 u + del p + f ]
=
[-((ux, uy, uz).d/dx ux + (ux, uy, uz).d/dy uy + (yx, uy, uz).d/dz uz) - v del^2 u + .... ]

Dot product itself is commutative, However, the case here is different as del 'operator' should be given a vector/magnitude to 'operate' on.


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Quote:
Original post by Darkstrike
The del.v notation is a formalism, you should not expect it to behave well.


This statement makes no sense.

To the OP: v is a scalar-valued function of x, y, and z. So is del.v.

v.del is a derivative operator, so it is not a scalar-valued function. This means del.v and v.del are not the same.

v.del is typically seen when computing directional derivatives of a function f(x,y,z). (v.del)f is the derivative of f in the (unit-length) direction v.

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Ah, thanks guys, now I understand...then, v.del is an derivative operator, and it still needs something on the right hand side to operate...Then, that means :

(u.del)u = u . del u = del u . u ?

Then, I think that it's correct to assume that the first term in Navier Stokes equation is :
-(u . del u)

However, to ensure incompressibility (divergence free) in Navier Stokes equation, we use this equation : del . u = 0, thus, still make the first term of Navier Stokes becomes zero...

[Edited by - EonStrife on September 30, 2007 2:32:41 AM]

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Quote:
Original post by Dave Eberly
Quote:
Original post by Darkstrike
The del.v notation is a formalism, you should not expect it to behave well.


This statement makes no sense.

del.v is not actually a dot product of a vector del and a vector v, so commutativity of the dot product does not imply that del.v=v.del. Does this sound better now?

edit:
Quote:
Original post by Dave Eberly
To the OP: v is a scalar-valued function of x, y, and z. So is del.v.

The OP's notation v = (vx, vy, vz) suggests that v is not a function, but rather a vector (or a 1-form, but then del v would be a two-form, not a function).

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Ah, I'm sorry for the equation above...
It should be : -(ux d/dx u + uy d/dy u + uz d/dz u) - v del^2 u + del p + f
as
-(u.del) = -[(ux d/dx) + (uy d/dy) + (uz d/dz)]

the rest is correct...;)

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@yehdev_cc
Hmm...ok, divergence works on velocity field...
Suppose given a velocity field : u = x^3yi + yzj+ xz^2i

So, div.u will become : d(x^3y)/dx + d(yz)/dy + d(xz^2)/dz = 3x^2y + z + 2xz

and about, -(u.div)u, from my understanding, it'll be like this:
http://img.photobucket.com/albums/v708/EonStrife/del.jpg

Am I correct? Thanks :)

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Quote:
Original post by Darkstrike
del.v is not actually a dot product of a vector del and a vector v, so commutativity of the dot product does not imply that del.v=v.del. Does this sound better now?


No. The dot product is used in a *formal* manner, so I would say that del.v is the dot product of a vector del and a vector v. In this manner, commutativity is not part of the discussion. The property of commutativity is associated with the definition of dot product of n-tuples whose components are scalar values.

Quote:

The OP's notation v = (vx, vy, vz) suggests that v is not a function, but rather a vector (or a 1-form, but then del v would be a two-form, not a function).


My error, and for hastily posting late at night. Yes, v is a vector-valued function (of a vector-valued variable).

del. is a differential operator that applies to vector fields v : R^n ->R^n. The result of application, del.v, is a scalar-valued function del.v : R^n -> R.

v.del is a differential operator that applies to scalar-valued functions u : R^n -> R. The result of application, (v.del)u, is a scalar-valued function (v.del)u : R^n -> R.

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Quote:
Original post by Dave Eberly
Quote:
Original post by Darkstrike
del.v is not actually a dot product of a vector del and a vector v, so commutativity of the dot product does not imply that del.v=v.del. Does this sound better now?


No. The dot product is used in a *formal* manner, so I would say that del.v is the dot product of a vector del and a vector v. In this manner, commutativity is not part of the discussion. The property of commutativity is associated with the definition of dot product of n-tuples whose components are scalar values.


No. One of the axioms which must be satisfied for an operator to be an inner product over a given vector space is commutativity. This holds for any vector space and is not specific to "n-tuples whose components are scalar".

The full list of axioms, for reference, are:

1. <x + y, z> = <x, z> + <y, z>
2. <x, y> = <y, x>
3. a <x, y> = x, y>
4. <x, x> >= 0 for all x
5. <x, x> = 0 iff x = 0

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