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Harmonic function squared and mean value

  1. Nov 7, 2013 #1
    1. The problem statement, all variables and given/known data
    Let u be a harmonic function in the open disk K centered at the origin with radius a. and
    [itex]∫_K[u(x,y)]^2 dxdy = M < ∞[/itex]. Prove that
    [itex]|u(x,y)| \le \frac{1}{a-\sqrt{x^2+y^2}}\left( \frac{M}{\pi}\right)^{1/2}[/itex] for all (x,y) in K.

    2. Relevant equations
    Mean value property for harmonic functions.

    3. The attempt at a solution
    I first thought this was easy and directly applied the mean property for harmonic functions, but of course, the square of a harmonic function is not harmonic (unless it's a constant).

    I can see this problems begs for the mean value using a ball with radius [itex]a-\sqrt{x^2+y^2}[/itex] which would be entirely inside the disk K but I can't get around the fact that u squared is not harmonic. Help please

    Attached Files:

    Last edited: Nov 7, 2013
  2. jcsd
  3. Nov 8, 2013 #2
    Are you integrating over the whole disk K (i.e. a double integral) or as a line integral around the boundary of K.
  4. Nov 8, 2013 #3
    I integrated over the area of the smaller circle inside with radius [itex] r=a-\sqrt{x^2+y^2}[/itex]. This is exactly what I did (I know it's wrong):

    [itex] u^2(x,y)=\frac{\int_{B_r(x,y)} u^2(w, z)dwdz}{\pi r^2}\leq \frac{\int_K u^2(w, z)dwdz}{\pi r^2}=\frac{M}{\pi r^2}\\
    |u(x,y)|\leq \sqrt{\frac{M}{\pi r^2}}[/itex]
    But of course, this would work if u^2 were harmonic, which is seldom the case.

    Any ideas? I was thinking maybe applying the maximum principle and somehow get bounds...
    Last edited: Nov 8, 2013
  5. Nov 9, 2013 #4
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