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Find harmonic conj of u(x,y)=ln(x^2+y^2)

  1. Jun 7, 2014 #1
    1. The problem statement, all variables and given/known data
    Find harmonic conjugate of u(x,y)=ln(x^2+y^2) and specify the region it is defined
    then show u has no harm conj on C\{0}


    2. Relevant equations



    3. The attempt at a solution
    Ok so i found the harmonic conj by converting to polar and found it to be v(r,Θ) = Θ.

    I am having trouble finding out where v is defined. Also it seems our analytic function f = u + iv
    is log(z) since u(r,Θ) = ln|r|. Isn't that defined everywhere but zero? Why would u not have a harmonic conj at C\{0}?
     
  2. jcsd
  3. Jun 7, 2014 #2

    mfb

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    The complex log can be defined everywhere but zero, but then you cannot make it continuous.
    The same problem comes up with v=Θ.
     
  4. Jun 7, 2014 #3
    If you dont mind, could you explain where the problem with v = Θ comes in?

    Also, since v = Θ is not continuously defined on C\{0} would that mean that it doesn't have a partial, and therefore cannot satisfy cauchy riemann, which in turn is the reason why u cannot have a harm conj on C\{0}?
     
  5. Jun 7, 2014 #4

    mfb

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    What happens if go around the unit circle, for example?

    I don't know what you mean here.
     
  6. Jun 7, 2014 #5
    Θ will go from 0 to 2pi


    I am trying to find the reason why v(r,Θ) not continuous on C\{0} implies u does not have a harmonic conjugate on C\{0}. Is it because if it is not continuous then it's partial derivative will not exist on C\{0}, thus implying that that Cauchy Riemann equations will not be satisfied on C\{0}?
     
  7. Jun 8, 2014 #6

    mfb

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    And what happens at the transition "close to 2 pi -> 0"?

    At least not everywhere. Right.
     
  8. Jun 8, 2014 #7
    When theta gets to 2pi it repeats. So at 0 and 2pi the function is multivalued thus not continuous?
     
  9. Jun 8, 2014 #8

    mfb

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    Right.
     
  10. Jun 8, 2014 #9
    Thanks for all the help these last couple of days mfb
     
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