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Prove the functions are unique in a volume, vector calculus problem

  1. Nov 13, 2012 #1
    1. The problem statement, all variables and given/known data
    In a volume V, enclosed by a surface S, the vector fields X and Y satisfy the coupled equations

    ∇×∇×X=X+Y

    ∇×∇×Y=YX

    If the values of ∇×X and ∇×Y are given on S, show that X and Y are unique in V.


    2. Relevant equations
    ∇.(A×B)=B.(∇×A)−A.(∇×B)

    ∇×(A×B)=∇(∇.A)−∇2A


    3. The attempt at a solution
    I am assuming that I need to show that ∇2X and ∇2Y are equal to zero to satisfy the uniqueness theorem for Poisson's equation. But am unsure of a good way to get there, so before I write my scribbles if someone could point me in the right direction it would be great.
     
  2. jcsd
  3. Nov 13, 2012 #2
    We need to prove two things, first that ∇ ^2 X = 0 and ∇ ^2 Y = 0, and also that X and Y are zero on S.

    It is straight forward to show that

    2X = ∇×∇×X - ∇×∇×Y
    2Y = ∇×∇×X + ∇×∇×Y

    Thus from the second identity above

    2X = ∇ (∇.X) - ∇^2 X - ∇ (∇.Y) + ∇^2
    2Y = ∇(∇.X) - ∇^2 X + ∇ (∇.Y) - ∇^2 Y

    Can we somehow now show from this using the divergence theorem that the above conditions are satisfied?
     
    Last edited: Nov 13, 2012
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