Let $u:U\to \mathbb{C}$ be harmonic and $\overline{D}(P,r)\subset U$. Verify the following variant of the mean value property of harmonic functions:

$$u(P)=\frac{1}{2\pi r}\int_{\partial D(P,r)}u(t)ds(t)$$ where $ds$ is arc length measure on $\partial D(P,r) \tag{$1$}$

I am not sure how to replace the $ds$ in $(1)$.

The mean value property states $$u(P)=\frac{1}{2\pi}\int_0 ^{2\pi}u(P+re^{i\theta})d\theta \tag{$2$}$$ so my idea is to start with the RHS of $(1)$ and to obtain the RHS of $(2)$. Using the substitution $t=P+re^{i\theta}$, the RHS of $(1)$ is $$\frac{1}{2\pi r}\int_{D(P,r)}u(t)ds(t)=\frac{1}{2\pi r}\int _0 ^{2\pi }u(P+re^{i\theta})ire^{i\theta}ds(\theta)\\=\frac{i}{2\pi}\int_0 ^{2\pi}u(P+re^{i\theta})e^{i\theta}ds(\theta) $$

How can I rewrite the $ds$ and how can I obtain the RHS of $(2)$?


1 Answer 1


This is really a notational mistake. A volume element $ds$ is just a one form on $\partial D$ so that $|ds| =1$. Now the boundary is parametrized by

$$ \gamma(\theta) = P + r e^{i\theta}\Rightarrow \gamma'(\theta) = ir e^{i\theta}\Rightarrow ||\gamma'(\theta)|| = r$$

Thus $ds = r d\theta$.

  • $\begingroup$ Thanks. I am not taking a norm or absolute value, so how do I get rid of the $ie^{i\theta}$ in my latest expression? $\endgroup$ Commented May 21, 2015 at 11:17
  • $\begingroup$ So what is your definition of $ds$? @TheSubstitute $\endgroup$
    – user99914
    Commented May 21, 2015 at 11:20
  • $\begingroup$ I am not quite sure. The problem states that it is arc length measure on $\partial D(P,r)$, but I am unsure of what that means. I don't see how this would differ if (1) had $dt$ instead of $ds(t)$ $\endgroup$ Commented May 21, 2015 at 11:22
  • 1
    $\begingroup$ In one variatble case, when we integrate a function $f$ on $[a, b]$, we always write $\int_a^b f(t) dt$. That because $||dt|| = 1$. But in general, $ds$ is given by $ds= |\gamma'(t)| dt$. @TheSubstitute $\endgroup$
    – user99914
    Commented May 21, 2015 at 11:25

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