# Integral Identity

1. May 28, 2010

### coverband

1. By considering, seperately, each component of the vector A, show that $$\iint A(u.n) ds = \iiint {(u.\nabla)A + A(\nabla.u)} dV$$ (A,u and n are vectors)

2. Relevant equations

3. Attempt at solution
L.H.S.
Let A = $$a\vec{i} + b\vec{j} + c\vec{k}$$
$$\iint (a\vec{i} + b\vec{j} + c\vec{k})[(u_1\vec{i} + u_2 \vec{j} + u_3 \vec{k}).(n_1 \vec{i} + n_2 \vec {j} + n_3 \vec{k})] ds = \iint (a\vec{i} + b\vec{j} + c\vec{k})(u_1 n_1 + u_2 n_2 + u_3 n_3) ds = \iint au_1n_1\vec{i} + au_2n_2\vec{i} + au_3n_3\vec{i} + bu_1n_1\vec{j}$$$$+ bu_2n_2\vec{j} + bu_3n_3\vec{j} + cu_1n_1\vec{k} + cu_2n_2\vec{k} + cu_3n_3\vec{k} ds$$

R.H.S.$$\iiint(u_1a_x\vec{i}+u_1b_x\vec{j}+u_1c_x\vec{k}+u_2a_y\vec{i}+u_2b_y\vec{j}+u_2c_y\vec{k}+u_3a_z\vec{i}+u_3b_z\vec{j}+u_3c_z\vec{k}+u_1a_x\vec{i}+u_1b_y\vec{i}+u_1c_z\vec{i}+u_2a_x\vec{j}+u_2b_y\vec{j}+u_2c_z\vec{j}+u_3a_x\vec{k}+u_3b_y\vec{k}+u_3c_z\vec{k})dv$$

Is this right? Where to now? Thanks

Last edited: May 28, 2010
2. May 28, 2010

### Jerbearrrrrr

Sum it up. i,j,k's are annoying D:!
WTS
$$\int A_i (u.n) ds$$
and
$$\int (u.\nabla) A_i +A_i (\nabla.u)dV$$
are the same.

We are treating A_i as a scalar function, for each i=1,2,3 separately.

Erm. Let me think what the best way to proceed is lol.

 Oh, I was right. You need to use a certain theorem that relates surface integrals to volume integrals. Note the dots and dels scattered around.
Haven't worked it all the way through but it should work.
[hint] $$\nabla . (a \vec {u})=$$?

Last edited: May 28, 2010
3. May 28, 2010

### coverband

How do you know to treat A as a scalar function?

With regard to what theorems encompass all above I would have to say the divergence theorem ?

4. May 28, 2010

### Jerbearrrrrr

Yes, div theorem.

For all intensive purposes, a vector is just n scalars in a bracket. (ordered tuple)

Consider a=b (a,b are vectors) - we can just treat this as n scalar equations a_i=b_i.

Also, the question says to consider separately the components of each bit. That means, equate the x component, the y component, the ... etc. And the equations you get are just scalar equations.

5. May 28, 2010

### coverband

Ok. So I just considered one of the vector A's components (say $$A_i$$).

Also I let the vector be equal to the product of a scalar function, B and a constant vector, a.

Thus equation becomes:

$$\iint (Ba u_1n_1 \vec{i} + Ba u_2n_2 \vec{i} + Ba u_3n_3 \vec{i}) ds$$$$= \iiint (u_1Ba_x \vec{i} + u_1B_xa \vec{i} + u_2Ba_y \vec{i} + u_2B_ya \vec{i} + u_3Ba_z \vec{i} + u_3B_za \vec {i} + u_1B_xa \vec {i}+ u_1B_xa \vec{i} + u_2B_xa \vec {i} + u_2Ba_x \vec {i} + u_3a_xB \vec {i} + u_3aB_x \vec {i}) dV$$

Little help!

6. May 28, 2010

### Jerbearrrrrr

Okay.
some days, you gotta expand all the crap out and check each little piece.

But today is not one of those days :p
Hopefully after this exercise, you'll be able to get a feel for the kind of thing you can do before resorting to expanding everything out completely.

Can you work out a nice expression for $$\nabla . (f \vec{u})$$? where f is a scalar function, and u is a vector-valued function.

7. May 28, 2010

### coverband

Well, I believe $$\nabla . (f \vec{u}) = f \nabla . (\vec{u}) + \nabla f . \vec{u}$$ which when $$\vec {u}$$ is constant reduces to $$\nabla . (f \vec{u}) = \nabla f . \vec{u}$$. How does this help....

8. May 28, 2010

### Jerbearrrrrr

yep. Usually this is written
$$\nabla .(fu) = f \nabla.u+(u.\nabla) f$$
which resembles the RHS of the expression we want (eg, take f as a component of A).

$$\int \nabla . (fu) dV = \int (fu).n dS$$ is the divergence theorem in this case. (u and n are vectors)
Now take f to be a component of A and see if you can relate this to what we want.
Note the alternate way of writing the dV integral, by using the "product rule" relation we just derived.

9. May 29, 2010

### coverband

Hi. Sorry for delay. I was trying to digest info.

So, 1. By considering, seperately, each component of the vector A, show that $$\iint A(u.n) ds = \iiint {(u.\nabla)A + A(\nabla.u)} dV$$ (A,u and n are vectors)

So R.H.S.
$$= \iiint a[(u.\nabla)f + f(\nabla.u)] dV = \iiint a{\nabla.(fu)} dV$$

L.H.S.
$$= \iint a(fu).n ds$$ (due to properties of dot product)

which is true because of the divergence theorem !? Is all that right? How do you cancel the "a's" ? Is fu a vector ? Thanks