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

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  • #1
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Homework Statement


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}


Homework Equations





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}?
 

Answers and Replies

  • #2
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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.
 
  • #3
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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.

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}?
 
  • #4
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If you dont mind, could you explain where the problem with v = Θ comes in?
What happens if go around the unit circle, for example?

Also, since v = Θ is not continuously defined on C\{0} would that mean that it doesn't have a partial
I don't know what you mean here.
 
  • #5
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What happens if go around the unit circle, for example?
Θ will go from 0 to 2pi


I don't know what you mean here.

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}?
 
  • #6
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Θ will go from 0 to 2pi
And what happens at the transition "close to 2 pi -> 0"?

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}?
At least not everywhere. Right.
 
  • #7
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When theta gets to 2pi it repeats. So at 0 and 2pi the function is multivalued thus not continuous?
 
  • #8
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When theta gets to 2pi it repeats. So at 0 and 2pi the function is multivalued thus not continuous?
Right.
 
  • #9
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Thanks for all the help these last couple of days mfb
 

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