Editing Divergence theorem (section)

==Proofs==
===For bounded open subsets of Euclidean space===
We are going to prove the following:{{citation needed|date=May 2024}}
{{math theorem|math_statement= 
Let <math>\Omega \subset \mathbb{R}^n</math> be open and bounded with <math>C^1</math> boundary. If <math>u</math> is <math>C^1</math> on an open neighborhood <math>O</math> of <math>\overline{\Omega}</math>, that is, <math>u \in C^1(O)</math>, then for each <math>i \in \{1, \dots, n\}</math>,
<math display="block">\int_{\Omega}u_{x_i}\,dV = \int_{\partial \Omega}u\nu_i\,dS,</math>
where <math>\nu : \partial \Omega \to \mathbb{R}^n </math> is the outward pointing unit normal vector to <math>\partial \Omega</math>.
Equivalently,
<math display="block">\int_{\Omega}\nabla u\,dV = \int_{\partial \Omega}u\nu\,dS.</math>
}}

'''Proof of Theorem.'''
<ref name="Alt 2016 p. ">{{cite book | last=Alt | first=Hans Wilhelm | title=Universitext | chapter=Linear Functional Analysis | publisher=Springer London | publication-place=London | year=2016 | isbn=978-1-4471-7279-6 | issn=0172-5939 | doi=10.1007/978-1-4471-7280-2 | pages=259–261, 270–272}}</ref>
{{ordered list
| 1 = The first step is to reduce to the case where <math>u \in C_c^1(\mathbb{R}^n)</math>. Pick <math>\phi \in C_c^{\infty}(O)</math> such that <math>\phi = 1</math> on <math>\overline{\Omega}</math>. Note that <math>\phi u \in C_c^{1}(O) \subset C_c^1(\mathbb{R}^n)</math> and <math>\phi u = u</math> on <math>\overline{\Omega}</math>. Hence it suffices to prove the theorem for <math>\phi u</math>. Hence we may assume that <math>u \in C_c^1(\mathbb{R}^n)</math>.

| 2 = Let <math>x_0 \in \partial \Omega</math> be arbitrary. The assumption that <math>\overline{\Omega}</math> has <math>C^1</math> boundary means that there is an open neighborhood <math>U</math> of <math>x_0</math> in <math>\mathbb{R}^n</math> such that <math>\partial \Omega \cap U</math> is the graph of a <math>C^1</math> function with <math>\Omega \cap U</math> lying on one side of this graph. More precisely, this means that after a translation and rotation of <math>\Omega</math>, there are <math>r > 0</math> and <math>h > 0</math> and a <math>C^1</math> function <math>g : \mathbb{R}^{n - 1} \to \mathbb{R}</math>, such that with the notation
<math display="block">x' = (x_1, \dots, x_{n - 1}),</math>
it holds that
<math display="block">U = \{x \in \mathbb{R}^n : |x'| < r \text{ and } |x_n - g(x')| < h\}</math>
and for <math>x \in U</math>,
<math display="block">
\begin{align}
    x_n = g(x') & \implies x \in \partial \Omega, \\
    -h < x_n - g(x') < 0 & \implies x \in \Omega, \\
    0 < x_n - g(x') < h & \implies x \notin \Omega. \\
\end{align}
</math>
Since <math>\partial \Omega</math> is compact, we can cover <math>\partial \Omega</math> with finitely many neighborhoods <math>U_1, \dots, U_N</math> of the above form. Note that <math>\{\Omega, U_1, \dots, U_N\}</math> is an open cover of <math>\overline{\Omega} = \Omega \cup \partial \Omega</math>. By using a <math>C^{\infty}</math> [[partition of unity]] subordinate to this cover, it suffices to prove the theorem in the case where either <math>u</math> has compact [[Support (mathematics)|support]] in <math>\Omega</math> or <math>u</math> has compact support in some <math>U_j</math>. If <math>u</math> has compact support in <math>\Omega</math>, then for all <math>i \in \{1, \dots, n\}</math>, <math>\int_{\Omega} u_{x_i}\,dV = \int_{\mathbb{R}^n}u_{x_i}\,dV = \int_{\mathbb{R}^{n - 1}} \int_{-\infty}^{\infty}u_{x_i}(x)\,dx_i\,dx' = 0</math> by the fundamental theorem of calculus, and <math>\int_{\partial \Omega}u\nu_i\,dS = 0</math> since <math>u</math> vanishes on a neighborhood of <math>\partial \Omega</math>. Thus the theorem holds for <math>u</math> with compact support in <math>\Omega</math>. Thus we have reduced to the case where <math>u</math> has compact support in some <math>U_j</math>.

| 3 = So assume <math>u</math> has compact support in some <math>U_j</math>. The last step now is to show that the theorem is true by direct computation. Change notation to <math>U = U_j</math>, and bring in the notation from (2) used to describe <math>U</math>. Note that this means that we have rotated and translated <math>\Omega</math>. This is a valid reduction since the theorem is invariant under rotations and translations of coordinates.  Since <math>u(x) = 0</math> for <math>|x'| \geq r</math> and for <math>|x_n - g(x')| \geq h</math>, we have for each <math>i \in \{1, \dots, n\}</math> that
<math display="block">
\begin{align}
    \int_{\Omega}u_{x_i}\,dV &= \int_{|x'| < r}\int_{g(x') - h}^{g(x')}u_{x_i}(x', x_n)\,dx_n\,dx' \\
    &= \int_{\mathbb{R}^{n - 1}}\int_{-\infty}^{g(x')}u_{x_i}(x', x_n)\,dx_n\,dx'.
\end{align}
</math>
For <math>i = n</math> we have by the fundamental theorem of calculus that
<math display="block">\int_{\mathbb{R}^{n - 1}}\int_{-\infty}^{g(x')}u_{x_n}(x', x_n)\,dx_n\,dx' = \int_{\mathbb{R}^{n - 1}}u(x', g(x'))\,dx'.</math>
Now fix <math>i \in \{1, \dots, n - 1\}</math>. Note that
<math display="block">\int_{\mathbb{R}^{n - 1}}\int_{-\infty}^{g(x')}u_{x_i}(x', x_n)\,dx_n\,dx' = \int_{\mathbb{R}^{n - 1}}\int_{-\infty}^{0}u_{x_i}(x', g(x') + s)\,ds\,dx'</math>
Define <math>v : \mathbb{R}^{n} \to \mathbb{R}</math> by <math>v(x', s) = u(x', g(x') + s)</math>. By the chain rule,
<math display="block">v_{x_i}(x', s) = u_{x_i}(x', g(x') + s) + u_{x_n}(x', g(x') + s)g_{x_i}(x').</math>
But since <math>v</math> has compact support, we can integrate out <math>dx_i</math> first to deduce that
<math display="block">\int_{\mathbb{R}^{n - 1}}\int_{-\infty}^{0}v_{x_i}(x', s)\,ds\,dx' = 0.</math>
Thus
<math display="block">
\begin{align}
    \int_{\mathbb{R}^{n - 1}}\int_{-\infty}^{0}u_{x_i}(x', g(x') + s)\,ds\,dx' &= \int_{\mathbb{R}^{n - 1}}\int_{-\infty}^{0}-u_{x_n}(x', g(x') + s)g_{x_i}(x')\,ds\,dx' \\
    &= \int_{\mathbb{R}^{n - 1}}-u(x', g(x'))g_{x_i}(x')\,dx'.
\end{align}
</math>
In summary, with <math>\nabla u = (u_{x_1}, \dots, u_{x_n})</math> we have
<math display="block">\int_{\Omega}\nabla u\,dV = \int_{\mathbb{R}^{n - 1}}\int_{-\infty}^{g(x')}\nabla u\,dV = \int_{\mathbb{R}^{n - 1}}u(x', g(x'))(-\nabla g(x'), 1)\,dx'.</math>
Recall that the outward unit normal to the graph <math>\Gamma</math> of <math>g</math> at a point <math>(x', g(x')) \in \Gamma</math> is <math>\nu(x', g(x')) = \frac{1}{\sqrt{1 + |\nabla g(x')|^2}}(-\nabla g(x'), 1)</math> and that the surface element <math>dS</math> is given by <math display="inline">dS = \sqrt{1 + |\nabla g(x')|^2}\,dx'</math>. Thus
<math display="block">\int_{\Omega}\nabla u\,dV  = \int_{\partial \Omega}u\nu\,dS.</math>
This completes the proof.
}}

===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.