Divergence free vector fields in R^n

[szf654]

Prove that every divergence free vector field on R^n, n>1 is of the form:

v(x)=SUM dAij/dxi *ej

where Aij(x) is smooth function from R^n to R such that Aij(x)=-Aji(x) i.e. matrix $[Aij(x)]$ is skew symmetric for every vector x.

 

[dextercioby]

Is this homework ? If so, you're in the wrong section and you should post your thoughts before asking for help. That's normally how homeworks are dealt with.

 

[szf654]

Thanks for pointing out, I am new here :)

 

[dextercioby]

OK, good. So what are your ideas to solve the problem ?

 

[szf654]

Oh, I reposted into homework section, but let me write here as well.


2. Homework Equations

Divergence of field v at p is Sum dv(x)/dxi. Vield is a divergence free if dv(x)/dxi=0 for every x.

Divergence theorem: integral div v over S=integral v*n over dS where n is vector pointing out on dS (boundary of S).

3. The Attempt at a Solution
First I checked that the converse is true, i.e. if the vector field v has form (**) it is divergence free. That can be done by simple calculation div v and using that Aij is smooth so d^2 Aij/dxi*dx=d^2 Aij / dxj*dxi (i.e. order of differentiation does not matter) and we also use Aij+Aji=0.

My attempt was to use induction as noted in the pset, i.e. by using divergence theorem we get that integral v*n over dS for every closed surface. My idea was to somewhat relate this integral to a div free vector field over R^{n-1}, but I did not manage that.

Second thing I tried is to somehow integrate over line parallel to e^j and get non differential relationship between v and Aij, but I am not sure if that is a legitimate field and how to define Aij in that way properly.

Thanks for help!

 

[dextercioby]

A_ij is the 2 form dual to a pseudovector in 3 D. Any pseudovector is the curl of a vector.

So

A_{ij} = \epsilon_{ijk} \tilde{B}_k = \epsilon_{ijk} (\nabla \times \vec{C})_k

Now you know that \nabla\cdot\vec{D} = 0. From these 2 things you can infer that

\vec{D} = \nabla \times\vec{C}

where \vec{C} is determined up to a gradient of a scalar.

 

[szf654]

Thanks, but to be honest I am familiar with the curl, but its the first time I hear about pseudovector and forms. Is there a way to avoid using it?

 

[dextercioby]

Divergence and curl are differential operators acting on vectors in R^3 and generally in R^n. Also the gradient and the laplacian. You must know about them, if you're into this problem.

 

[szf654]

Well in my class we did not define them, just the divergence, and we are still required to solve this problem.

 

[dextercioby]

In this case, you can exploit the fact that the derivative operators commute and contract them with an arbitrary antisymmetric tensor.

\partial_i v_i = 0 \leftrightarrow v_i = \partial_j A_{ij}, A_{ij} = -A_{ji}

To show that such a tensor exists means to solve a system of PDEs of 1st order (3eqns for 3 independent components of A).

 

[szf654]

It seems that your look at this problem is way above my level of understanding :)