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Divergence free vector fields in R^n

  1. Dec 6, 2011 #1
    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.
     
  2. jcsd
  3. Dec 6, 2011 #2

    dextercioby

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    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.
     
  4. Dec 6, 2011 #3
    Thanks for pointing out, I am new here :)
     
  5. Dec 6, 2011 #4

    dextercioby

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    OK, good. So what are your ideas to solve the problem ?
     
  6. Dec 6, 2011 #5
    Oh, I reposted into homework section, but let me write here as well.


    2. Relevant 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 paralel 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!
     
  7. Dec 6, 2011 #6

    dextercioby

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

    So

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

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

    [tex] \vec{D} = \nabla \times\vec{C} [/tex]

    where [itex] \vec{C}[/itex] is determined up to a gradient of a scalar.
     
    Last edited: Dec 6, 2011
  8. Dec 6, 2011 #7
    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?
     
  9. Dec 6, 2011 #8

    dextercioby

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    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.
     
  10. Dec 6, 2011 #9
    Well in my class we did not define them, just the divergence, and we are still required to solve this problem.
     
  11. Dec 6, 2011 #10

    dextercioby

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

    [tex] \partial_i v_i = 0 \leftrightarrow v_i = \partial_j A_{ij} [/tex], [itex] A_{ij} = -A_{ji} [/itex]

    To show that such a tensor exists means to solve a system of PDEs of 1st order (3eqns for 3 independent components of A).
     
  12. Dec 6, 2011 #11
    It seems that your look at this problem is way above my level of understanding :)
     
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