N-dimensional cube problem with clever solution?

[isison]

Homework Statement

[PLAIN]http://img202.imageshack.us/img202/4535/asdty.jpg

Homework Equations

I think the relevant equations are the divergence theorem
[URL]http://upload.wikimedia.org/math/0/9/b/09b08625d572f4d7c9ceb7585241dc36.png[/URL]

The Attempt at a Solution

Since it hints at a "clever" solution, I assume the difficult part is to find the right path. I think we can take advantage of the divergence theorem to convert the surface integral to a volume integral (well, it's n-dimensional, so from n-1 -> n).
I think the flux on opposite sides are canceled out, because the function is a function of radius only? But the velocity can be different on the opposite sides.

 

[eok20]

What did you get when you converted the surface integral to a volume integral? What can you say about \nabla \cdot (\nabla u)?

 

[isison]

What did you get when you converted the surface integral to a volume integral? What can you say about \nabla \cdot (\nabla u)?

Thanks for the reply.

I see that by using the divergence theorem, we can convert it into a laplace equation, and since u is harmonic, that term is zero. However, how do we take the velocity into account? I'm confused about that.

 

[Dick]

Thanks for the reply.

I see that by using the divergence theorem, we can convert it into a laplace equation, and since u is harmonic, that term is zero. However, how do we take the velocity into account? I'm confused about that.

\nu is not a velocity. It's the outward pointing normal vector. And you can't apply the divergence theorem to the cube since the divergence is undefined at the origin. Use the divergence theorem to argue that you change the integration surface from a cube to a sphere. Then do the surface integration over the sphere instead.

 

[isison]

\nu is not a velocity. It's the outward pointing normal vector. And you can't apply the divergence theorem to the cube since the divergence is undefined at the origin. Use the divergence theorem to argue that you change the integration surface from a cube to a sphere. Then do the surface integration over the sphere instead.

Thanks for the input.

Please help me understand this: I see this question with great similarity with the usual gauss' theorem problem I did in EE classes. I don't see how you can argue that the cubic surface can be reduced to a sphere by Divergence theorem; integrating the cubic surface and sphere contains different volumn, thus different amount of charge.

The v from what I understand is a vector, but I don't think it's a unit vector. It could vary with position and be different at each surface.

Can you please explain to me a bit more of your prospective on this problem?

 

[Dick]

It's essentially the same problem you got in your EE class except more dimensions. By v being a unit vector, they mean it has unit length and it's normal to the surface. The point about the divergence is that since you've shown u is harmonic, use that it's harmonic to show the divergence of grad(u) is zero. That means the change in volume between cube and sphere doesn't matter.