Establishing a Variant of the Mean Value Property of Harmonic Functions

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 $$(2)$$?

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 $ds = r d\theta$.
• 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? – The Substitute May 21 '15 at 11:17
• So what is your definition of $ds$? @TheSubstitute – user99914 May 21 '15 at 11:20
• 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 $ds(t)$ – The Substitute May 21 '15 at 11:22
• 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 – user99914 May 21 '15 at 11:25