# An integral estimation in the proof of the uniqueness of Neumann problem for $\triangle$

This is probably a follow-up question to this one. In the sequel, let $D\subset{\mathbb R}^m(m=2,3)$ be a bounded domain of class $C^2$. And assume that the boundary $\partial D$ is connected. [EDIT: By $\nu$ we denote the unit normal of $\partial D$ directed into the exterior domain ${\mathbb R}^m\setminus\bar D$.]

Exterior Neumann Problem. Find a function $u$ that is harmonic in ${\mathbb R}^m\setminus \bar D$, is continuous in ${\mathbb R}^m\setminus D$, and satisfies the boundary condition $$\frac{\partial u}{\partial \nu} = g\qquad \text{on}~\partial D,$$ in the sense $$\lim_{h\to+0}\nu(x)\cdot\nabla u(x+h\nu(x))=g(x),\quad x\in\partial D$$ of uniform convergence on $\partial D$, where $g$ is a given continuous function. For $|x|\to\infty$ it is required that $u(x)=o(1)$ uniformly for all directions.

For understanding of the proof of uniqueness of the Exterior Neumann Problem, here is my question:

How can I get $$\lim_{r\to\infty}\int_{\Omega_r}u\frac{\partial u}{\partial \nu}ds = 0$$ and $$\lim_{h\to 0 }\int_{\partial D_h}u\frac{\partial u}{\partial \nu}ds = 0$$ where $u$ is the solution to the Exterior Neumann Problem where $g=0$, $\Omega_r$ is the sphere of radius $r$, $\partial D_h:=\{x+h\nu(x):x\in\partial D\}$ with sufficiently small $h>0$?

Intuitively these two are correct since $u(x)=o(1)$, as $|x|\to\infty$ and $\frac{\partial u}{\partial \nu}=0$ on $\partial D$ and a KNOWN fact that $$\nabla u(x)=O\bigg(\frac{1}{|x|^{m-1}}\bigg),\quad |x|\to \infty$$

But I don't know how to write down the rigorous proof.

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Is $\nu$ assumed to be perpendicular to $\partial D$? –  robjohn Sep 10 '11 at 2:34
@robjohn: Edited. Thanks. –  Jack Sep 10 '11 at 2:45
Using Green's first identity I get $$-\int_{\mathbb{R}^m}\|\nabla u\|^2dx$$ for the first limit. Did you forget a factor or am I not understanding the measure you're using? For the second integral, show that $\sup\|\partial_v u\|\to0$ on $\partial D_h$ as $h\to0$. –  anon Sep 10 '11 at 4:40
@anon: I think the integral I'm using is the surface integral. –  Jack Sep 10 '11 at 13:18