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Let $\Omega\subset\mathbb R^n$ be an limited open set, and the boundary $\partial\Omega$ connected, and $u:\mathbb R^n\rightarrow \mathbb R$ a $C^\infty$ function, and $n$ the normal vector to $\partial \Omega$, then $$ \int_\Omega \Delta u=\int_{\partial\Omega} \langle\nabla u, n\rangle $$

I've tried to prove using Stokes Theorem, taking $d\omega=\sum\frac{\partial^2u}{\partial x_i^2}dx_1\wedge\dots\wedge dx_n$, we can se that $\omega=\sum (-1)^{i+1}\frac{\partial u}{\partial dx_i}dx_1\wedge\dots\wedge\hat{dx_i}\wedge\dots\wedge dx_n$, so I got $$ \int_\Omega \Delta u=\int_{ \Omega}d\omega=\int_{\partial\Omega}\omega=\int_{\partial\Omega}\sum (-1)^{i+1}\frac{\partial u}{\partial dx_i}dx_1\wedge\dots\wedge\hat{dx_i}\wedge\dots\wedge dx_n, $$ doing the pull-back I got $$ \int_{\partial\Omega}(-1)^{i+1}\frac{\partial u}{\partial x_i}\det\begin{bmatrix} \frac{\partial\varphi_1}{dy_1}&\dots&\frac{\partial\varphi_1}{dy_{n-1}}\\ &\vdots&\\ \hat{\frac{\partial\varphi_i}{dy_1}}&\dots&\hat{\frac{\partial\varphi_i}{dy_{n-1}}}\\ &\vdots&\\ \frac{\partial\varphi_n}{dy_1}&\dots&\frac{\partial\varphi_n}{dy_{n-1}} \end{bmatrix}dy, $$ where $\varphi=(\varphi_1,\dots,\varphi_n)$ is a parametrization of $\partial\Omega$, so I know that when this determinatn is in $\mathbb R^3$ it is exactly the vector product (with a correction of sign that cancels the $(-1)^{i+1}$), and then result follows, but does this result follows in $\mathbb R^n$, i.e, this determinat is $(-1)^{i+1}\frac{\partial\varphi}{dy_1}\times\dots\times\hat{\frac{\partial\varphi}{dy_{i}}}\dots\times\frac{\partial\varphi}{dy_{n-1}}$?

Thanks in advance.

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  • $\begingroup$ You are on the right track. The following may be helpful: There is always an isomorphism $\phi: U\subset \mathbb{R}^n \to V \subset \mathbb{R}^n$, such that the restriction $\phi|_{V\times\{0\}}$ for $V\times\{0\} \subset U$ is a local parametrization of $\partial \Omega$ and $\frac{\partial}{\partial y_n} \phi = n$. $\endgroup$
    – mlk
    Jun 16, 2017 at 13:55
  • $\begingroup$ Should it really be $\frac{\partial u}{\partial dx_i}$, not $\frac{\partial u}{\partial x_i}$ in the formulas? Or is it some notation that I don't know of? $\endgroup$
    – md2perpe
    Jun 16, 2017 at 16:20

1 Answer 1

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Let $\upsilon = dx^1 \wedge \cdots \wedge dx^n$ and $\sigma_k = (-1)^k dx^1 \wedge \cdots \widehat{dx^k} \cdots \wedge dx^n$. Let $F^k$ be the $k$-component of a vector field $F$. Then $$ \int_{\partial\Omega} F \cdot n = \int_{\partial\Omega} F^k \sigma_k = \int_{\Omega} d(F^k \sigma_k) = \int_{\Omega} \partial_j F^k dx^j \wedge \sigma_k \\ = \int_{\Omega} \partial_j F^k \delta_k^j \upsilon = \int_{\Omega} \partial_j F^j \upsilon = \int_{\Omega} \nabla \cdot F $$

Taking $F = \nabla u$ gives $$ \int_{\partial\Omega} \langle\nabla u, n\rangle = \int_\Omega \nabla \cdot \nabla u = \int_\Omega \Delta u $$

I understand that it should be shown (if it hasn't already in the course) that $\sigma_k = n_k \sigma$ where $\sigma$ is the area form on $\partial \Omega$.

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