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Complex analysis harmonic function

  1. May 30, 2009 #1
    I seem to be missing a subtlety of the definition of a harmonic function. I'm using Churchill and Brown. As stated in the book, an analytic function in domain D with component functions (i.e. real and imaginary parts) u(x,y) and v(x,y) are harmonic in D.

    harmonic functions satisfy uxx+uyy=0

    Consider u=x3 +y

    then uxx+uyy gives 6x+0 [tex]\neq[/tex] 0

    but this is analytic in some domain D.

    So what am I missing here, why does this not satisfy the conditions to be a harmonic function?
     
  2. jcsd
  3. May 30, 2009 #2

    Mute

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    How can you claim it's analytic if it doesn't satisfy Laplace's Equation? The fact that u is not harmonic should imply that u(x,y) = x^3 + y is not the real part of any analytic function, no?
     
  4. May 30, 2009 #3
    I claim it's analytic because it is differentiable in some domain D. Which it clearly is, it's just a polynomial which is differentiable.
     
  5. May 30, 2009 #4

    mathman

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    As a real function of two variables, u is analytic. As the real part of a complex function, it is not.
     
  6. May 30, 2009 #5
    hm... can you say why it is not intuitively without saying "it does not satidfy la place's eqn?
     
  7. May 30, 2009 #6
    Solved, found someone from my class, Cauchy-Riemann equations do not hold.
     
  8. May 30, 2009 #7

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    Note that the analyticity referred to in complex analysis is the analyticity of the complex function, f(z). It doesn't refer to the "real function" analyticity of u or v.

    If f(z) = u(x,y) + i*v(x,y) is an analytic function, then u and v must satisfy the Cauchy-Riemann equations, which in turn imply they must each satisfy the 2d Laplace Equation.

    I can't immediately think of any "intuitive" way to look at some arbitrary u(x,y) and deduce it is not the real part of an analytic function without just doing the check to see if it satisfies Laplace's equation.
     
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