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Let $A \subset \mathbb{R}^2$ be open and connected and let $u \in C^2(A)$ be harmonic. Then $u$ satisfies $$u(x)=\frac{1}{2\pi}\int_0^{2\pi}u(x+r\hat n(\theta)) \, d\theta$$

I'm given the following proof:

Let $x \in A$ and take $r>0$ so that $B_r(x) \subset A$ . Since $u$ is harmonic it satisfies $$0= \int_{B_r(x)} \Delta u(y) \, dy = \int_{B_r(0)} \Delta u(x+y) \, dy = \int_{\partial B_r(0)} \nabla u(x+y) \cdot \hat{n} \, ds$$ and then some more follows from here.

My problem is with how the last equality is made. I know that the divergence theorem says that $$\int_V \nabla\cdot F \, dV = \int_{\partial V} F \cdot d\vec S = \int_{\partial V}F \cdot \hat{n} \, dS$$ where $V$ is some volume in $\mathbb{R}^3$.

So is some sort of analagous statement in $\mathbb{R}^2$ being used here? That is, is there a statement saying that $$\int_S \nabla \cdot F \, dS = \int_{\partial S} F \cdot d\vec{r}$$? And then I assume $d\vec{r} = \hat{n} \, ds$

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Yes, the divergence theorem holds in any dimension (even in one dimension where it's known as the fundamental theorem of calculus).

There are two approaches how to formulate it, either you define the surface integral of a scalar or you define the directed surface integral of a field both being equivalent:

$$\int_{\partial S} u \overline{dS} = \int_{\partial S} u\cdot \overline{n}dS$$

Perhaps you by $\overrightarrow{dr}$ mean the (normal) directed surface measure?

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  • $\begingroup$ Thank you. Am I correct in saying that $d\vec{r} = \hat{n} \, ds$? $\endgroup$
    – MHW
    Jun 29 '17 at 13:16
  • $\begingroup$ @Si.0788: Not exactly. The correct statement of the divergence theorem in the plane is $\int_S \nabla \cdot F\,dA = \int_{\partial S} F \cdot \hat n\,ds$. I believe $d\vec r$ is usually an abbreviation for $\hat t \,ds$. $\endgroup$ Jun 29 '17 at 13:19
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Using the Divergence Theorem, $$ \begin{align} 0 &=\frac1r\iint_{B(c,r)}\Delta u\,\mathrm{d}x\,\mathrm{d}y\\ &=\frac1r\int_{\partial B(c,r)}n\cdot\nabla u\,\mathrm{d}s\\ &=\int_{\partial B(c,r)}n\cdot\nabla u\,\frac{\mathrm{d}s}r\\ &=\int_{\partial B(c,r)}n\cdot\nabla u\,\mathrm{d}\theta\\ &=\frac{\partial}{\partial r}\int_{\partial B(c,r)}u\,\mathrm{d}\theta \end{align} $$ Then note that $$ \lim_{r\to0}\int_{\partial B(c,r)}u\,\mathrm{d}\theta=u(c) $$

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