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Harmonic functions

  1. Feb 11, 2009 #1
    the function [tex]f:\Re^n \rightarrow \Re = |X|^{2-n} [/tex] is harmonic, but only for n > 2.

    why is the condition n > 2 imposed. Isn't harmonic for all values of n??
     
  2. jcsd
  3. Feb 11, 2009 #2
    Shouldn't the domain be [tex]\mathbb{R}^n\setminus \{ 0 \}[/tex]? Otherwise, said f could not even exist at 0, and hence is clearly not harmonic.
     
  4. Feb 12, 2009 #3
    ya i'm sorry that is the domain...my mistake...

    but still, why shoud n>2, why cant n = 1???..

    note: n stands for dimension of the space and hence is an integer >=1
     
  5. Feb 12, 2009 #4

    malawi_glenn

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    oh, is this classical physics? ;-)

    Try what happens if n=1, is the function harmonic then?
     
  6. Feb 12, 2009 #5
    of course it is classical physics. It plays a MAJOR role in electromagnetic theory. and the potential formulation of the theory.

    for n = 1, it is harmonic as laplacian(|X|) = 0 right? so it satisfies.

    but then, why put the condition n > 2. I have seen it in many books
     
  7. Feb 12, 2009 #6

    malawi_glenn

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    well then calculus problems are classic physics problems aswell. Just trying to make sure you get help.

    Now harmonic functions are functions which are continuously differentiable twice and satisfy Laplace equation.

    So play around with the criterion for twice continuously differentiable.
     
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