User:IlIIllIIIIlllIlIIllllIllllII/sandbox

Theorem — Let ${\overline {\Omega }}$ be a $C^{2}$ compact manifold with boundary with $C^{1}$ metric tensor $g$. Let $\Omega$ denote the manifold interior of ${\overline {\Omega }}$ and let $\partial \Omega$ denote the manifold boundary of ${\overline {\Omega }}$. Let $(\cdot ,\cdot )$ denote $L^{2}$ inner products of functions and $\langle \cdot ,\cdot \rangle$ denote inner products of vectors. Suppose $u\in C^{1}({\overline {\Omega }},\mathbb {R} )$ and $X$ is a $C^{1}$ vector field on ${\overline {\Omega }}$. Then

$$(\operatorname {grad} u,X)=-(u,\operatorname {div} X)+\int _{\partial \Omega }u\langle X,N\rangle \,dS,$$

where $N$ is the outward-pointing unit normal vector to $\partial \Omega$.

Proof of Theorem. We use the Einstein summation convention. By using a partition of unity, we may assume that $u$ and $X$ have compact support in a coordinate patch $O\subset {\overline {\Omega }}$. First consider the case where the patch is disjoint from $\partial \Omega$. Then $O$ is identified with an open subset of $\mathbb {R} ^{n}$ and integration by parts produces no boundary terms:

$${\begin{aligned}(\operatorname {grad} u,X)&=\int _{O}\langle \operatorname {grad} u,X\rangle {\sqrt {g}}\,dx\\&=\int _{O}\partial _{j}uX^{j}{\sqrt {g}}\,dx\\&=-\int _{O}u\partial _{j}({\sqrt {g}}X^{j})\,dx\\&=-\int _{O}u{\frac {1}{\sqrt {g}}}\partial _{j}({\sqrt {g}}X^{j}){\sqrt {g}}\,dx\\&=(u,-{\frac {1}{\sqrt {g}}}\partial _{j}({\sqrt {g}}X^{j}))\\&=(u,-\operatorname {div} X).\end{aligned}}$$

In the last equality we used the Voss-Weyl coordinate formula for the divergence, although the preceding identity could be used to define $-\operatorname {div}$ as the formal adjoint of $\operatorname {grad}$. Now suppose $O$ intersects $\partial \Omega$. Then $O$ is identified with an open set in $\mathbb {R} _{+}^{n}=\{x\in \mathbb {R} ^{n}:x_{n}\geq 0\}$. We zero extend $u$ and $X$ to $\mathbb {R} _{+}^{n}$ and perform integration by parts to obtain

$${\begin{aligned}(\operatorname {grad} u,X)&=\int _{O}\langle \operatorname {grad} u,X\rangle {\sqrt {g}}\,dx\\&=\int _{\mathbb {R} _{+}^{n}}\partial _{j}uX^{j}{\sqrt {g}}\,dx\\&=(u,-\operatorname {div} X)-\int _{\mathbb {R} ^{n-1}}u(x',0)X^{n}(x',0){\sqrt {g(x',0)}}\,dx',\end{aligned}}$$

where $dx'=dx_{1}\dots dx_{n-1}$. By a variant of the straightening theorem for vector fields, we may choose $O$ so that ${\frac {\partial }{\partial x_{n}}}$ is the inward unit normal $-N$ at $\partial \Omega$. In this case ${\sqrt {g(x',0)}}\,dx'={\sqrt {g_{\partial \Omega }(x')}}\,dx'=dS$ is the volume element on $\partial \Omega$ and the above formula reads

$$(\operatorname {grad} u,X)=(u,-\operatorname {div} X)+\int _{\partial \Omega }u\langle X,N\rangle \,dS.$$

This completes the proof.