# Help with a Derivative

1. Oct 24, 2009

I am working on deriving an acceleration field. We have the velocity field

$$\mathbf{V}(\mathbf{r},t) = \mathbf{i}u(x,y,z,t) + \mathbf{j}v(x,y,z,t) + \mathbf{k}w(x,y,z,t)\,\,\,\,\,\,\,(1)$$

where u,v, and w are the scalar components Vx, Vy, and Vz, respectively.

$$\therefore\,\,\,\,\,\,\,\,\,\mathbf{a}=\frac{d\,\mathbf{V}}{d\,t} = \mathbf{i}\frac{d\,u}{d\,t} + \mathbf{j}\frac{d\,v}{d\,t} + \mathbf{i}\frac{d\,w}{d\,t} \,\,\,\,\,\,\,(2)$$

Looking at just the first term, we have by the chain rule:

$$\frac{d\,u(x,y,z,t)}{d\,t} = \frac{\partial{u}}{\partial{t}} + \frac{\partial{u}}{\partial{x}}\frac{d\,x}{d\,t} + \frac{\partial{u}}{\partial{y}}\frac{d\,y}{d\,t} + \frac{\partial{u}}{\partial{z}}\frac{d\,z}{d\,t} \,\,\,\,\,\,\,(3)$$

But since dx/dt = u, dy/dt = v and dz/dt = w, we can write:

$$\frac{d\,u(x,y,z,t)}{d\,t} = \frac{\partial{u}}{\partial{t}} + u\frac{\partial{u}}{\partial{x}} + v\frac{\partial{u}}{\partial{y}} + w\frac{\partial{u}}{\partial{z}} \,\,\,\,\,\,\,\,(4)$$

Here is where I get major confused. My text denotes this as:

$$\mathbf{a} = \frac{\partial{u}}{\partial{t}} + (\mathbf{V}\cdot\nabla)u\,\,\,\,\,\,\,\,(5)$$

I get that

$$(\mathbf{V}\cdot\nabla) = \partial{\mathbf{V_x}/\partial{x} + \partial{\mathbf{V_y}}/\partial{y} +\partial{\mathbf{V_z}}/\partial{z} = \partial{\mathbf{u}/\partial{x} + \partial{\mathbf{v}}/\partial{y} +\partial{\mathbf{w}}/\partial{z} \,\,\,\,\,\,\,\,(6)$$

I don't understand how multiplying EQ (6) by 'u' gets you the last 3 terms in EQ (4) ?

What about 'v' and 'w' ?

Where am I getting confused?

Is my EQ (6) right?

Last edited: Oct 24, 2009
2. Oct 24, 2009

### LCKurtz

You have the expression ut +uux + vuy + wuz which is ut + $<u, v, w> \cdot <u_x,u_y,u_z>$, which is just $u_t + V \cdot \nabla u$.

And $V \cdot \nabla u$ is exactly what $(V \cdot \nabla)u$ means. The $\nabla$ operator only affects terms to its right.

3. Oct 24, 2009

Okay. So you are saying that it is not actually being 'multiplied by u' ? Instead, 'u' is being operated on by nabla ?

Also, does this mean that $\mathbf{V}\cdot\nabla\ne\nabla\cdot\mathbf{V}$?

4. Oct 24, 2009

### LCKurtz

Yes. That is exactly right.

5. Oct 24, 2009

Oh wow. I was thinking that $\mathbf{V}\cdot\nabla=\nabla\cdot\mathbf{V}$ and was taking the divergence of V-->$\nabla\cdot\mathbf{V}$ I am still a little confused.

But I think I can possibly work through it now. Why do they notate it $(V \cdot \nabla)u$ ?? Shouldn't it be $\mathbf{V} \cdot (\mathbf{\nabla u})$?

The former doesn't even make any sense How can you 'dot' a Vector with an operator?

And moreover, if you are going to 'dot' something that doesn't make any sense then it should have to be commutative by definition of dot product.

Anyone have any thoughts on this? I could use some clarity here.

Last edited: Oct 24, 2009
6. Oct 24, 2009

### LCKurtz

The nabla operator is sometimes written an an operator form like this:

$$\nabla = i\frac{\partial}{\partial x} + j\frac{\partial}{\partial y}+ k\frac{\partial}{\partial z}$$

or in other notation:

$$\langle\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\rangle$$

It operates on vectors with the usual dot and cross properties much like the D differentiation operator:

(D2 + 2D + 1)y = y'' + 2y' + 1

If $V = \langle a,b,c \rangle$ then

$$V \cdot \nabla = a\frac{\partial}{\partial x} + b\frac{\partial}{\partial y}+ c\frac{\partial}{\partial z}$$

and it is ready to "multiply", that is operate, on a scalar on its right. Does that help? And, yes, you lose the commutative property of dot product with nabla.

7. Oct 24, 2009