Incompressible Navier Stokes - Short Question

1. Jan 2, 2010

Kastenfrosch

Hello!

The incompressible Navier Stokes equation consists of the two equations

and

Why can't i insert the 2nd one into the first one so that the advection term drops out?!
$$\nabla\cdot$$v = v$$\cdot\nabla$$ = 0
=>
(v$$\cdot\nabla)\cdot$$v = 0

Last edited by a moderator: Apr 24, 2017
2. Jan 2, 2010

Mapes

Hi Kastenfrosch, welcome to PF. $\nabla\cdot \bold{v}$ is not a dot product but is rather shorthand for

$$\frac{\partial v_x}{\partial x}+\frac{\partial v_y}{\partial y}+\frac{\partial v_z}{\partial z}$$

There is no such single term as $\bold{v}\cdot\nabla$. (This dual use of the dot notation has caused a lot of confusion; see, for example, Tai's "http://deepblue.lib.umich.edu/bitstream/2027.42/7869/5/bad1475.0001.001.pdf"").

Last edited by a moderator: Apr 24, 2017
3. Jan 4, 2010

Kastenfrosch

Hello and Thanks for your answer!

... but i think i still dont know what to do....

in the linked PDF i saw that there are many definitions, but i didn't find a definition for $$v \cdot\nabla$$

encouraged by your post i searched for "abuse of nabla", and i found that it's not right to always treat $$\nabla$$ as a vector of partial derivatives

But if i can't treat

$$v \cdot \nabla$$

as

$$\left(v_x,v_y,v_z\right)^T \cdot \left( \frac{d}{dx}, \frac{d}{dy}, \frac{d}{dz} \right)^T$$

, how can i calculate it?

in
http://en.wikipedia.org/wiki/Advection#Mathematics_of_advection
they treat $$v \cdot \nabla$$ just as the commutative inner product, don't they?!

Last edited: Jan 4, 2010
4. Jan 4, 2010

Mapes

I don't see how it's a product; there's no such entity as "$\nabla$." It's an operator:

$$(\bold{v}\cdot\nabla)\bold{v}=\left(v_x\frac{\partial v_x}{\partial x}+v_y\frac{\partial v_x}{\partial y}+v_z\frac{\partial v_x}{\partial z}\right)\bold{i}+\left(v_x\frac{\partial v_y}{\partial x}+v_y\frac{\partial v_y}{\partial y}+v_z\frac{\partial v_y}{\partial z}\right)\bold{j}+\left(v_x\frac{\partial v_z}{\partial x}+v_y\frac{\partial v_z}{\partial y}+v_z\frac{\partial v_z}{\partial z}\right)\bold{k}$$

which is not related to $\nabla\cdot\bold{v}$!

5. Jan 4, 2010

Kastenfrosch

Sorry, perhaps i get you wrong because i'm from germany...

So

$$(\nabla \cdot \bold{v})\cdot \bold{v} = \left( \frac{\partial v_x}{\partial x} + \frac{\partial v_y}{\partial y} + \frac{\partial v_z}{\partial z} \right) \bold{v}$$
=
$$\left(v_x\frac{\partial v_x}{\partial x}+v_x\frac{\partial v_y}{\partial y}+v_x\frac{\partial v_z}{\partial z}\right)\bold{i}+\left(v_y\frac{\partial v_x}{\partial x}+v_y\frac{\partial v_y}{\partial y}+v_y\frac{\partial v_z}{\partial z}\right)\bold{j}+\left(v_z\frac{\partial v_x}{\partial x}+v_z\frac{\partial v_y}{\partial y}+v_z\frac{\partial v_z}{\partial z}\right)\bold{k}$$

and so
$$(\nabla \cdot \bold{v}) \cdot \bold{v} \neq (\bold{v} \cdot \nabla) \bold{v}$$
(because your big-tearm was another one)

But how did you know how $$(\bold{v}\cdot\nabla)\bold{v}$$ is computed?
Do you first resolve $$\bold{v} \cdot \nabla$$ and afterwards multiply with $$\bold{v}$$?

And why do they use inner-product-notation when they don't treat nabla as the partial-derivatives vector and the dot as the dot-product?

Last edited: Jan 5, 2010
6. Jan 4, 2010

Mapes

I had to look it up.

Because it can be convenient (though risky). (See http://en.wikipedia.org/wiki/Abuse_of_notation#Del_operator" for an explanation.)

Last edited by a moderator: Apr 24, 2017
7. Jan 4, 2010

tiny-tim

Welcome to PF!

Hi Kastenfrosch! Welcome to PF!

(have a del: ∇ )

If v is a vector, then there's no such thing as ∇v (because ∇ without a dot or a cross can only act on a scalar).

So there really isn't anything else that (v.∇)v could mean.

8. Jan 5, 2010

Brian_C

Actually, I have seen the gradient operator applied to a vector before. You just take the gradient of each vector component and add them up vectorially. We know that the vector gradient is not used in this case because of the parentheses.

9. Jan 5, 2010

tiny-tim

Hi Brian_C!
But that's not a vector.

If we call that ∇B, then ∇B(f(x,y),0) = (∂f/∂x,∂f/∂y).

Now rotate the coordinates by a fixed angle θ …

B(f(x,y)cosθ,f(x,y)sinθ) = (∂f/∂x,∂f/∂y)(cosθ + sinθ), which is a completely different vector.

10. Jan 5, 2010

Brian_C

You are right. I was thinking of the Laplacian operator (del squared).

11. Jan 5, 2010

tiny-tim

dodgy

Yup … I think that was the dodgy Peckham operator (del trotter).

12. Jan 5, 2010

Kastenfrosch

Ah, ok, if i got you right, you mean that

$$(\nabla \cdot \bold{v}) = \left( \frac{\partial v_x}{\partial x} + \frac{\partial v_y}{\partial y} + \frac{\partial v_z}{\partial z} \right)$$

whereas

$$(\bold{v} \cdot \nabla) = \left( \frac{v_x \partial }{\partial x} + \frac{v_y \partial }{\partial y} + \frac{v_z \partial }{\partial z} \right)$$

which is not the same, so

$$(\bold{v} \cdot \nabla) \neq (\nabla \cdot \bold{v})$$

so i can't substitute one for another.

If i'm right (please give me a short feedback), i'm really gratefull for having that much patience with me :).

If i'm wrong: buhuuu :(

13. Jan 5, 2010

tiny-tim

Hi Kastenfrosch!

Yes and no …

Yes $$(\nabla \cdot \bold{v}) = \left( \frac{\partial v_x}{\partial x} + \frac{\partial v_y}{\partial y} + \frac{\partial v_z}{\partial z} \right)$$

but

$$(\bold{v} \cdot \nabla) = \left( \frac{v_x \partial }{\partial x},\ \frac{v_y \partial }{\partial y},\ \frac{v_z \partial }{\partial z} \right)$$

So the first one is a scalar operating on a vector, but the second one is a vector operating on a vector.

Your second one was a scalar, operating on a scalar, the dot-product of v with the gradient: v.(∇f)

14. Jan 5, 2010

Kastenfrosch

OK, then i still didn't get it....

in http://en.wikipedia.org/wiki/Advection
they say, that $$\bold{v}\cdot\nabla$$ is a scalar.

And if i use
$$(\bold{v} \cdot \nabla) = \left( \frac{v_x \partial }{\partial x} + \frac{v_y \partial }{\partial y} + \frac{v_z \partial }{\partial z} \right)$$
and multiply it with v to get
$$(\bold{v} \cdot \nabla) \cdot \bold{v}$$
i get the same result as Mapes in his second Post

So was it your mistake or am i standing on the hosepipe (german expression for temporarily not understanding obvious things)?

Last edited: Jan 5, 2010
15. Jan 5, 2010

tiny-tim

Yes, but that is "the first one" (a scalar operating on a vector).

And it's an unsatisfactory way of writing it.
No you don't, it's a scalar (operating on a vector), and if you "multiply" it by v you get

$$(\bold{v} \cdot \nabla)\bold{v} = \left( \frac{v_x \partial\bold{v}}{\partial x} + \frac{v_y \partiall\bold{v}}{\partial y} + \frac{v_z \partiall\bold{v}}{\partial z} \right)$$

(And you can't "dot" it with v, because it isn't written as a vector, and you can only "dot" two vectors.

But Mapes's result is right … I got confused by the absence of the second v (but it's still a vector operating on a vector)

16. Jan 6, 2010

Kastenfrosch

what do you mean with "the first one" and "the last one"?
1. = $$(\nabla \cdot \bold{v})\bold{v}$$
2. = $$(\bold{v} \cdot \nabla)\bold{v}$$
?

i can see my mistake, that i wrote a $$\cdot$$ between $$(\nabla \cdot \bold{v})$$ and $$\bold{v}$$, which is no dot- but a scalar multiplication... But what i meant was multiplying the scalar with the vector.

So if i exchange my dot with a scalar-multiplication sign, i'm right with my last two posts?

17. Jan 6, 2010

tiny-tim

"The first one" as in …
(and I didn't say "the last one" )
No, your …
is still wrong, it's a scalar (with "+"s), and you need a vector (with ","s).

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