Integral Identity: Showing LHS = RHS

[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)

Homework Equations

3. Attempt at solution
L.H.S.
Let A = a\vec{i} + b\vec{j} + c\vec{k}
<br /> \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<br /> <br /> = \iint (a\vec{i} + b\vec{j} + c\vec{k})(u_1 n_1 + u_2 n_2 + u_3 n_3) ds<br /> <br /> = \iint au_1n_1\vec{i} + au_2n_2\vec{i} + au_3n_3\vec{i} + bu_1n_1\vec{j}<br /> + 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.<br /> \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

 

[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.

[edit] 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})=?

 

[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 ?

 

[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.

 

[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<br /> <br /> = <br /> \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!

 

[Jerbearrrrrr]

Okay.
some days, you got to 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.

 

[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...

 

[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.

 

[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<br /> <br /> = \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