Divergence theorem

We are going to prove the following:

Theorem — Let ${\displaystyle \Omega \subset \mathbb {R} ^{n}}$ be open and bounded with ${\displaystyle C^{1}}$ boundary. If ${\displaystyle u}$ is ${\displaystyle C^{1}}$ on an open neighborhood ${\displaystyle O}$ of ${\displaystyle {\overline {\Omega }}}$, that is, ${\displaystyle u\in C^{1}(O)}$, then for each ${\displaystyle i\in \{1,\dots ,n\}}$,

$${\displaystyle \int _{\Omega }u_{x_{i}}\,dV=\int _{\partial \Omega }u\nu _{i}\,dS,}$$

where ${\displaystyle \nu$ :\partial \Omega \to \mathbb {R} ^{n}} is the outward pointing unit normal vector to ${\displaystyle \partial \Omega }$. Equivalently,

$${\displaystyle \int _{\Omega }\nabla u\,dV=\int _{\partial \Omega }u\nu \,dS.}$$

Proof of Theorem. [6] (1) The first step is to reduce to the case where ${\displaystyle u\in C_{c}^{1}(\mathbb {R} ^{n})}$. Pick ${\displaystyle \phi \in C_{c}^{\infty }(O)}$ such that ${\displaystyle \phi =1}$ on ${\displaystyle {\overline {\Omega }}}$. Note that ${\displaystyle \phi u\in C_{c}^{1}(O)\subset C_{c}^{1}(\mathbb {R} ^{n})}$ and ${\displaystyle \phi u=u}$ on ${\displaystyle {\overline {\Omega }}}$. Hence it suffices to prove the theorem for ${\displaystyle \phi u}$. Hence we may assume that ${\displaystyle u\in C_{c}^{1}(\mathbb {R} ^{n})}$.

(2) Let ${\displaystyle x_{0}\in \partial \Omega }$ be arbitrary. The assumption that ${\displaystyle {\overline {\Omega }}}$ has ${\displaystyle C^{1}}$ boundary means that there is an open neighborhood ${\displaystyle U}$ of ${\displaystyle x_{0}}$ in ${\displaystyle \mathbb {R} ^{n}}$ such that ${\displaystyle \partial \Omega \cap U}$ is the graph of a ${\displaystyle C^{1}}$ function with ${\displaystyle \Omega \cap U}$ lying on one side of this graph. More precisely, this means that after a translation and rotation of ${\displaystyle \Omega }$, there are ${\displaystyle r>0}$ and ${\displaystyle h>0}$ and a ${\displaystyle C^{1}}$ function ${\displaystyle g:\mathbb {R} ^{n-1}\to \mathbb {R} }$, such that with the notation

$${\displaystyle x'=(x_{1},\dots ,x_{n-1}),}$$

it holds that

$${\displaystyle U=\{x\in \mathbb {R} ^{n}:|x'|<r{\text{ and }}|x_{n}-g(x')|<h\}}$$

and for ${\displaystyle x\in U}$,

$${\displaystyle {\begin{aligned}x_{n}=g(x')&\implies x\in \partial \Omega ,\\-h<x_{n}-g(x')<0&\implies x\in \Omega ,\\0<x_{n}-g(x')<h&\implies x\notin \Omega .\\\end{aligned}}}$$

Since ${\displaystyle \partial \Omega }$ is compact, we can cover ${\displaystyle \partial \Omega }$ with finitely many neighborhoods ${\displaystyle U_{1},\dots ,U_{N}}$ of the above form. Note that ${\displaystyle \{\Omega ,U_{1},\dots ,U_{N}\}}$ is an open cover of ${\displaystyle {\overline {\Omega }}=\Omega \cup \partial \Omega }$. By using a ${\displaystyle C^{\infty }}$ partition of unity subordinate to this cover, it suffices to prove the theorem in the case where either ${\displaystyle u}$ has compact support in ${\displaystyle \Omega }$ or ${\displaystyle u}$ has compact support in some ${\displaystyle U_{j}}$. If ${\displaystyle u}$ has compact support in ${\displaystyle \Omega }$, then for all ${\displaystyle i\in \{1,\dots ,n\}}$, ${\displaystyle \int _{\Omega }u_{x_{i}}\,dV=\int _{\mathbb {R} ^{n}}u_{x_{i}}\,dV=\int _{\mathbb {R} ^{n-1}}\int _{-\infty }^{\infty }u_{x_{i}}(x)\,dx_{i}\,dx'=0}$ by the fundamental theorem of calculus, and ${\displaystyle \int _{\partial \Omega }u\nu _{i}\,dS=0}$ since ${\displaystyle u}$ vanishes on a neighborhood of ${\displaystyle \partial \Omega }$. Thus the theorem holds for ${\displaystyle u}$ with compact support in ${\displaystyle \Omega }$. Thus we have reduced to the case where ${\displaystyle u}$ has compact support in some ${\displaystyle U_{j}}$.

(3) So assume ${\displaystyle u}$ has compact support in some ${\displaystyle U_{j}}$. The last step now is to show that the theorem is true by direct computation. Change notation to ${\displaystyle U=U_{j}}$, and bring in the notation from (2) used to describe ${\displaystyle U}$. Note that this means that we have rotated and translated ${\displaystyle \Omega }$. This is a valid reduction since the theorem is invariant under rotations and translations of coordinates. Since ${\displaystyle u(x)=0}$ for ${\displaystyle |x'|\geq r}$ and for ${\displaystyle |x_{n}-g(x')|\geq h}$, we have for each ${\displaystyle i\in \{1,\dots ,n\}}$ that

$${\displaystyle {\begin{aligned}\int _{\Omega }u_{x_{i}}\,dV&=\int _{|x'|<r}\int _{g(x')-h}^{g(x')}u_{x_{i}}(x',x_{n})\,dx_{n}\,dx'\\&=\int _{\mathbb {R} ^{n-1}}\int _{-\infty }^{g(x')}u_{x_{i}}(x',x_{n})\,dx_{n}\,dx'.\end{aligned}}}$$

For ${\displaystyle i=n}$ we have by the fundamental theorem of calculus that

$${\displaystyle \int _{\mathbb {R} ^{n-1}}\int _{-\infty }^{g(x')}u_{x_{n}}(x',x_{n})\,dx_{n}\,dx'=\int _{\mathbb {R} ^{n-1}}u(x',g(x'))\,dx'.}$$

Now fix ${\displaystyle i\in \{1,\dots ,n-1\}}$. Note that

$${\displaystyle \int _{\mathbb {R} ^{n-1}}\int _{-\infty }^{g(x')}u_{x_{i}}(x',x_{n})\,dx_{n}\,dx'=\int _{\mathbb {R} ^{n-1}}\int _{-\infty }^{0}u_{x_{i}}(x',g(x')+s)\,ds\,dx'}$$

Define ${\displaystyle v:\mathbb {R} ^{n}\to \mathbb {R} }$ by ${\displaystyle v(x',s)=u(x',g(x')+s)}$. By the chain rule,

$${\displaystyle v_{x_{i}}(x',s)=u_{x_{i}}(x',g(x')+s)+u_{x_{n}}(x',g(x')+s)g_{x_{i}}(x').}$$

But since ${\displaystyle v}$ has compact support, we can integrate out ${\displaystyle dx_{i}}$ first to deduce that

$${\displaystyle \int _{\mathbb {R} ^{n-1}}\int _{-\infty }^{0}v_{x_{i}}(x',s)\,ds\,dx'=0.}$$

Thus

$${\displaystyle {\begin{aligned}\int _{\mathbb {R} ^{n-1}}\int _{-\infty }^{0}u_{x_{i}}(x',g(x')+s)\,ds\,dx'&=\int _{\mathbb {R} ^{n-1}}\int _{-\infty }^{0}-u_{x_{n}}(x',g(x')+s)g_{x_{i}}(x')\,ds\,dx'\\&=\int _{\mathbb {R} ^{n-1}}-u(x',g(x'))g_{x_{i}}(x')\,dx'.\end{aligned}}}$$

In summary, with ${\displaystyle \nabla u=(u_{x_{1}},\dots ,u_{x_{n}})}$ we have

$${\displaystyle \int _{\Omega }\nabla u\,dV=\int _{\mathbb {R} ^{n-1}}\int _{-\infty }^{g(x')}\nabla u\,dV=\int _{\mathbb {R} ^{n-1}}u(x',g(x'))(-\nabla g(x'),1)\,dx'.}$$

Recall that the outward unit normal to the graph ${\displaystyle \Gamma }$ of ${\displaystyle g}$ at a point ${\displaystyle (x',g(x'))\in \Gamma }$ is ${\displaystyle \nu (x',g(x'))={\frac {1}{\sqrt {1+|\nabla g(x')|^{2}}}}(-\nabla g(x'),1)}$ and that the surface element ${\displaystyle dS}$ is given by ${\displaystyle dS={\sqrt {1+|\nabla g(x')|^{2}}}\,dx'}$. Thus

$${\displaystyle \int _{\Omega }\nabla u\,dV=\int _{\partial \Omega }u\nu \,dS.}$$

This completes the proof.

We are going to prove the following:

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

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

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

Proof of Theorem. [7] We use the Einstein summation convention. By using a partition of unity, we may assume that ${\displaystyle u}$ and ${\displaystyle X}$ have compact support in a coordinate patch ${\displaystyle O\subset {\overline {\Omega }}}$. First consider the case where the patch is disjoint from ${\displaystyle \partial \Omega }$. Then ${\displaystyle O}$ is identified with an open subset of ${\displaystyle \mathbb {R} ^{n}}$ and integration by parts produces no boundary terms:

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

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

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

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

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

This completes the proof.

As a result of the divergence theorem, a host of physical laws can be written in both a differential form (where one quantity is the divergence of another) and an integral form (where the flux of one quantity through a closed surface is equal to another quantity). Three examples are Gauss's law (in electrostatics), Gauss's law for magnetism, and Gauss's law for gravity.

Continuity equations offer more examples of laws with both differential and integral forms, related to each other by the divergence theorem. In fluid dynamics, electromagnetism, quantum mechanics, relativity theory, and a number of other fields, there are continuity equations that describe the conservation of mass, momentum, energy, probability, or other quantities. Generically, these equations state that the divergence of the flow of the conserved quantity is equal to the distribution of sources or sinks of that quantity. The divergence theorem states that any such continuity equation can be written in a differential form (in terms of a divergence) and an integral form (in terms of a flux).[12]

Any inverse-square law can instead be written in a Gauss's law-type form (with a differential and integral form, as described above). Two examples are Gauss's law (in electrostatics), which follows from the inverse-square Coulomb's law, and Gauss's law for gravity, which follows from the inverse-square Newton's law of universal gravitation. The derivation of the Gauss's law-type equation from the inverse-square formulation or vice versa is exactly the same in both cases; see either of those articles for details.[12]

To verify the planar variant of the divergence theorem for a region ${\displaystyle R}$:

${\displaystyle R=\left\{(x,y)\in \mathbb {R} ^{2}\$ :\ x^{2}+y^{2}\leq 1\right\},}

and the vector field:

${\displaystyle \mathbf {F} (x,y)=2y\mathbf {i} +5x\mathbf {j} .}$

The boundary of ${\displaystyle R}$ is the unit circle, ${\displaystyle C}$, that can be represented parametrically by:

${\displaystyle x=\cos(s),\quad y=\sin(s)}$

such that ${\displaystyle 0\leq s\leq 2\pi }$ where ${\displaystyle s}$ units is the length arc from the point ${\displaystyle s=0}$ to the point ${\displaystyle P}$ on ${\displaystyle C}$. Then a vector equation of ${\displaystyle C}$ is

${\displaystyle C(s)=\cos(s)\mathbf {i} +\sin(s)\mathbf {j} .}$

At a point ${\displaystyle P}$ on ${\displaystyle C}$:

${\displaystyle P=(\cos(s),\,\sin(s))\,\Rightarrow \,\mathbf {F} =2\sin(s)\mathbf {i} +5\cos(s)\mathbf {j} .}$

Therefore,

${\displaystyle {\begin{aligned}\oint _{C}\mathbf {F} \cdot \mathbf {n} \,\mathrm {d} s&=\int _{0}^{2\pi }(2\sin(s)\mathbf {i} +5\cos(s)\mathbf {j} )\cdot (\cos(s)\mathbf {i} +\sin(s)\mathbf {j} )\,\mathrm {d} s\\&=\int _{0}^{2\pi }(2\sin(s)\cos(s)+5\sin(s)\cos(s))\,\mathrm {d} s\\&=7\int _{0}^{2\pi }\sin(s)\cos(s)\,\mathrm {d} s\\&=0.\end{aligned}}}$

Because ${\displaystyle M=2y}$, we can evaluate ${\displaystyle {\frac {\partial M}{\partial x}}=0}$, and because ${\displaystyle N=5x}$, ${\displaystyle {\frac {\partial N}{\partial y}}=0}$. Thus

${\displaystyle \iint _{R}\,\mathbf {\nabla } \cdot \mathbf {F} \,\mathrm {d} A=\iint _{R}\left({\frac {\partial M}{\partial x}}+{\frac {\partial N}{\partial y}}\right)\,\mathrm {d} A=0.}$

Let's say we wanted to evaluate the flux of the following vector field defined by ${\displaystyle \mathbf {F} =2x^{2}{\textbf {i}}+2y^{2}{\textbf {j}}+2z^{2}{\textbf {k}}}$ bounded by the following inequalities:

${\displaystyle \left\{0\leq x\leq 3\right\}\left\{-2\leq y\leq 2\right\}\left\{0\leq z\leq 2\pi \right\}}$

By the divergence theorem,

${\displaystyle \iiint _{V}\left(\mathbf {\nabla } \cdot \mathbf {F} \right)\mathrm {d} V=}$ ${\displaystyle \scriptstyle S}$ ${\displaystyle (\mathbf {F} \cdot \mathbf {n} )\,\mathrm {d} S.}$

We now need to determine the divergence of ${\displaystyle {\textbf {F}}}$. If ${\displaystyle \mathbf {F} }$ is a three-dimensional vector field, then the divergence of ${\displaystyle {\textbf {F}}}$ is given by ${\textstyle \nabla \cdot {\textbf {F}}=\left({\frac {\partial }{\partial x}}{\textbf {i}}+{\frac {\partial }{\partial y}}{\textbf {j}}+{\frac {\partial }{\partial z}}{\textbf {k}}\right)\cdot {\textbf {F}}}$.

Thus, we can set up the following flux integral ${\displaystyle I=}$ ${\displaystyle {\scriptstyle S}}$ ${\displaystyle \mathbf {F} \cdot \mathbf {n} \,\mathrm {d} S,}$ as follows:

${\displaystyle {\begin{aligned}I&=\iiint _{V}\nabla \cdot \mathbf {F} \,\mathrm {d} V\\[6pt]&=\iiint _{V}\left({\frac {\partial \mathbf {F_{x}} }{\partial x}}+{\frac {\partial \mathbf {F_{y}} }{\partial y}}+{\frac {\partial \mathbf {F_{z}} }{\partial z}}\right)\mathrm {d} V\\[6pt]&=\iiint _{V}(4x+4y+4z)\,\mathrm {d} V\\[6pt]&=\int _{0}^{3}\int _{-2}^{2}\int _{0}^{2\pi }(4x+4y+4z)\,\mathrm {d} V\end{aligned}}}$

Now that we have set up the integral, we can evaluate it.

${\displaystyle {\begin{aligned}\int _{0}^{3}\int _{-2}^{2}\int _{0}^{2\pi }(4x+4y+4z)\,\mathrm {d} V&=\int _{-2}^{2}\int _{0}^{2\pi }(12y+12z+18)\,\mathrm {d} y\,\mathrm {d} z\\[6pt]&=\int _{0}^{2\pi }24(2z+3)\,\mathrm {d} z\\[6pt]&=48\pi (2\pi +3)\end{aligned}}}$

One can use the general Stokes' Theorem to equate the n-dimensional volume integral of the divergence of a vector field F over a region U to the (n − 1)-dimensional surface integral of F over the boundary of U:

${\displaystyle \underbrace {\int \cdots \int _{U}} _{n}\nabla \cdot \mathbf {F} \,\mathrm {d} V=\underbrace {\oint _{}\cdots \oint _{\partial U}} _{n-1}\mathbf {F} \cdot \mathbf {n} \,\mathrm {d} S}$

This equation is also known as the divergence theorem.

When n = 2, this is equivalent to Green's theorem.

When n = 1, it reduces to the fundamental theorem of calculus, part 2.

Writing the theorem in Einstein notation:

${\displaystyle \iiint _{V}{\dfrac {\partial \mathbf {F} _{i}}{\partial x_{i}}}\mathrm {d} V=}$ ${\displaystyle \scriptstyle S}$ ${\displaystyle \mathbf {F} _{i}n_{i}\,\mathrm {d} S}$

suggestively, replacing the vector field F with a rank-n tensor field T, this can be generalized to:[20]

${\displaystyle \iiint _{V}{\dfrac {\partial T_{i_{1}i_{2}\cdots i_{q}\cdots i_{n}}}{\partial x_{i_{q}}}}\mathrm {d} V=}$ ${\displaystyle \scriptstyle S}$ ${\displaystyle T_{i_{1}i_{2}\cdots i_{q}\cdots i_{n}}n_{i_{q}}\,\mathrm {d} S.}$

where on each side, tensor contraction occurs for at least one index. This form of the theorem is still in 3d, each index takes values 1, 2, and 3. It can be generalized further still to higher (or lower) dimensions (for example to 4d spacetime in general relativity[21]).