Editing Divergence theorem (section)

===For compact Riemannian manifolds with boundary===
We are going to prove the following:{{citation needed|date=May 2024}}
{{math theorem|math_statement=
Let <math>\overline{\Omega}</math> be a <math>C^2</math> compact manifold with boundary with <math>C^1</math> metric tensor <math>g</math>. Let <math>\Omega</math> denote the manifold interior of <math>\overline{\Omega}</math> and let <math>\partial \Omega</math> denote the manifold boundary of <math>\overline{\Omega}</math>. Let <math>(\cdot, \cdot)</math> denote <math>L^2(\overline{\Omega})</math> inner products of functions and <math>\langle \cdot, \cdot \rangle</math> denote inner products of vectors. Suppose <math>u \in C^{1}(\overline{\Omega}, \mathbb{R})</math> and <math>X</math> is a <math>C^1</math> vector field on <math>\overline{\Omega}</math>. Then
<math display="block">
(\operatorname{grad} u, X) = -(u, \operatorname{div} X) + \int_{\partial \Omega}u\langle X, N \rangle\,dS,
</math>
where <math>N</math> is the outward-pointing unit normal vector to <math>\partial \Omega</math>.
}}

'''Proof of Theorem.'''
<ref name="Taylor 2011 p. ">
{{cite book
|last=Taylor
|first=Michael E.
|title=Applied Mathematical Sciences
|chapter=Partial Differential Equations I
|publisher=Springer New York
|publication-place=New York, NY
|year=2011
|volume=115
|isbn=978-1-4419-7054-1
|issn=0066-5452
|doi=10.1007/978-1-4419-7055-8
|pages=178–179}}
</ref>
We use the Einstein summation convention. By using a partition of unity, we may assume that <math>u</math> and <math>X</math> have compact support in a coordinate patch <math>O \subset \overline{\Omega}</math>. First consider the case where the patch is disjoint from <math>\partial \Omega</math>. Then <math>O</math> is identified with an open subset of <math>\mathbb{R}^n</math> and integration by parts produces no boundary terms:
<math display="block">
\begin{align}
    (\operatorname{grad} u, X) &= \int_{O}\langle \operatorname{grad} u, X \rangle \sqrt{g}\,dx \\
    &= \int_{O}\partial_j u X^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{align}
</math>
In the last equality we used the Voss-Weyl coordinate formula for the divergence, although the preceding identity could be used to define <math>-\operatorname{div}</math> as the formal adjoint of <math>\operatorname{grad}</math>. Now suppose <math>O</math> intersects <math>\partial \Omega</math>. Then <math>O</math> is identified with an open set in <math>\mathbb{R}_{+}^n = \{x \in \mathbb{R}^n : x_n \geq 0\}</math>. We zero extend <math>u</math> and <math>X</math> to <math>\mathbb{R}_+^n</math> and perform integration by parts to obtain
<math display="block">
\begin{align}
    (\operatorname{grad} u, X) &= \int_{O}\langle \operatorname{grad} u, X \rangle \sqrt{g}\,dx \\
    &= \int_{\mathbb{R}_+^n}\partial_j u X^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{align}
</math>
where <math>dx' = dx_1 \dots dx_{n - 1}</math>.
By a variant of the [[straightening theorem for vector fields]], we may choose <math>O</math> so that <math>\frac{\partial}{\partial x_n}</math> is the inward unit normal <math>-N</math> at <math>\partial \Omega</math>. In this case <math>\sqrt{g(x', 0)}\,dx' = \sqrt{g_{\partial \Omega}(x')}\,dx' = dS</math> is the volume element on <math>\partial \Omega</math> and the above formula reads
<math display="block">
(\operatorname{grad} u, X) = (u, -\operatorname{div} X) + \int_{\partial \Omega}u\langle X, N \rangle \,dS.
</math>
This completes the proof.