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Homework Help: Harmonic Function Values (Complex Analysis)

  1. Jun 26, 2012 #1
    1. The problem statement, all variables and given/known data
    Let [itex]u[/itex] be a continuous real-valued function in the closure of the unit disk [itex]\mathbb{D}[/itex] that is harmonic in [itex]\mathbb{D}[/itex]. Assume that the boundary values of [itex]u[/itex] are given by

    [itex] u(e^{it}) = 5- 4 \cos t. [/itex]

    Furthermore, let [itex]v[/itex] be a harmonic conjugate of [itex]u[/itex] in [itex]\mathbb{D}[/itex] such that [itex]v(0) = 1[/itex]. Find [itex]u(1/2)[/itex] and [itex]v(1/2)[/itex].

    2. Relevant equations

    Harmonic functions fulfill
    [itex] u_{xx} + u_{yy} = 0[/itex].
    Poisson's Integral Formula (for harmonic functions) tells us that
    [itex] u(z) = \frac{1}{2\pi} \int_{0}^{2\pi} \frac{1-|z|^2}{|e^{i\theta}-z|^2}u(e^{i\theta}) d\theta [/itex].
    The harmonic conjugate of [itex]u[/itex] is given by the line integral
    [itex] v(z) = \operatorname{Im} \int_0 ^z u_x(w) - iu_y(w) dw [/itex].
    The harmonic conjugate is itself harmonic, so Poisson's formula applies to it as well.

    3. The attempt at a solution

    It's easy to find [itex]u(1/2)[/itex]: it is
    [itex] \frac{1}{2\pi} \int_0^{2\pi} \frac{3/4}{5/4 - \cos\theta} (5-4\cos \theta) d\theta = \frac{1}{2\pi} \int_0 ^{2\pi} 3 d\theta = 3 [/itex].

    However, I can't figure out how to compute [itex]v(1/2)[/itex]; the formulas just seem to give a hopelessly complicated mess.
    Any ideas on how to proceed?
  2. jcsd
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