I can't seem to find any references that gives a proof of the Gauss-Green theorem:

Let $U\subset\mathbb{R}^{n}$ be an open, bounded set with $\partial U$ being $C^1$. Suppose $u\in C^{1}(\bar{U})$, then $$\int_{U}{\frac{\partial u}{\partial x_i}}dx=\int_{\partial U}u\nu^{i}dS\;\;\;\;(i=1,\ldots,n),$$ where $\nu=(\nu^1,\ldots\nu^n)$ denotes the outward-pointing unit normal vector field to the region $U$.

Evans' PDE textbook presents the theorem (with no proof) in the appendix, and proceeds to use it to derive Green's formulas and the formula for $n$-dimensional integration by parts. So I'd really like to have a proof of the theorem for future reference.

  • $\begingroup$ Does anyone have any purely-analytic proof? I'm really not very comfortable with my geometry. $\endgroup$
    – Patch
    Feb 29, 2012 at 21:07

4 Answers 4


There is a simple proof of Gauss-Green theorem if one begins with the assumption of Divergence theorem, which is familiar from vector calculus, \begin{equation} \int_{U}\mathrm{div}\,\mathbf{w}\,dx = \int_{\partial U} \mathbf{w}\cdot\mathbf{\nu}\,dS, \end{equation} where $\mathbf{w}$ is any $C^\infty$ vector field on $U\in\Bbb{R}^n$ and $\mathbf{\nu}$ is the outward normal on $\partial U$.

Now, given the scalar function $u$ on the open set $U$, we can construct the vector field \begin{equation} \mathbf{w}=(0,\ldots,0,u,0,\ldots,0), \end{equation} where $u$ is the $i$th component. Then, following the Divergence theorem, we have \begin{equation} \int_U \mathrm{div}\,\mathbf{w}\,dx=\int_U u_{x_i}\,dx =\int_{\partial U}\mathbf{w}\cdot\mathbf{\nu}\,dS =\int_{\partial U}u\nu^i\,dS. \end{equation}

In Evans' book (Page 712), the Gauss-Green theorem is stated without proof and the Divergence theorem is shown as a consequence of it. This may be opposite to what most people are familiar with.


Here is an analytic proof not using differential forms. Restricted to $n=3$, it comes almost verbatim from Stewart, Calculus: Early Transcendentals.

To get started, let us naively assume that region $U$ is convex. The reason for us to make the seemingly too-much assumption is that we want to give a description of $U$ in terms of a single two sided inequality (in the form $a\leq x_i\leq b$) for each $i$. Let $D$ be the projection of $U$ onto the $x_1x_2\cdots\hat{x_i}\cdots x_n$ hyperplane (to avoid confusion, $\hat{}$ means the variable under it is absent). Due to the convexity of $U$, we can write $U=\{(x_1,\cdots,x_n)\in \mathbf{R}^n|(x_1,\cdots,\hat{x_i},\cdots,x_n)\in D\mbox{ and } f_1\leq x_i \leq f_2\}$ for some $\mathbf{R}^{n-1}\to \mathbf{R}$ functions $f_1(x_1,\cdots,\hat{x_i},\cdots,x_n)$ and $f_2(x_1,\cdots,\hat{x_i},\cdots,x_n)$. Here $f_1$ and $f_2$ are $C^1$ by the definition of $\partial U$ being $C^1$(see the appendix in Evans).

Remember we want to prove $$\int_U\frac{\partial u}{\partial x_i}dx=\int_{\partial U}u\nu^idS.$$ The description of $U$ above allows us to write the LHS as an iterated integral $$\int_D\left(\int_{f_1(x_1,\cdots,\hat{x_i},\cdots,x_n)}^{f_2(x_1,\cdots,\hat{x_i},\cdots,x_n)}\frac{\partial u}{\partial x_i}dx_i\right)dA,$$ where $dA=dx_1\cdots\hat{dx_i}\cdots dx_n$. Apply the Fundamental Theorem of Calculus to the inner integral, we then have $$\int_Du(x_1,\cdots,f_2(x_1,\cdots,\hat{x_i},\cdots,x_n),\cdots,x_n)-u(x_1,\cdots,f_1(x_1,\cdots,\hat{x_i},\cdots,x_n)\cdots,x_n)dA.$$

This is all we can do to the LHS now. For the RHS, notice that $\partial U$ can be decomposed into three surfaces $S_2$, $S_3$ and $S_{1}$ where $\nu^i$ at all the points of $S_2$ are positive, $\nu^i$ at all the points of $S_3$ are zero ($S_3$ is parallel to the $x_i$ axis) and $\nu^i$ of $S_1$ are negative. Thus, RHS is


Since $\nu_i$ for $S_3$ is zero, only the first and last term keep. From the geometric picture described at the beginning, the projection of $S_1$ and $S_2$ onto the $x_1x_2\cdots\hat{x_i}\cdots x_n$ hyperplane are exactly $D$. Also keep in mind that $\nu^i$ is the direction cosine of $\nu$ with the $x_i$ axis in $\mathbf{R}^n$. Therefore, $$\int_{S_2}u\nu^idS=\int_Du(x_1,\cdots,f_2(x_1,\cdots,\hat{x_i},\cdots,x_n),\cdots,x_n)d A,$$ and $$\int_{S_1}u\nu^idS=-\int_Du(x_1,\cdots,f_1(x_1,\cdots,\hat{x_i},\cdots,x_n),\cdots,x_n)dA.$$

This proves the theorem when $U$ is convex. For general $C^1$ $U$, I am thinking to cut the region into convex chunks so that when we glue them together the surface integral over touching faces are just cancelled due to opposite normal directions. Of course, I hope occasional "non-smoothie" won't introduce too much trouble.

  • $\begingroup$ Could you send please the page and the edition of the Stewart book where you get the reference? $\endgroup$
    – JuanMuñoz
    May 20, 2020 at 12:40
  • $\begingroup$ just an observation: The proof is basically equivalent to pairing up elements in an infinite summation to get another expression for the same summation. $\endgroup$ Mar 7, 2021 at 7:52

It is a special case of both Stokes' theorem, and the Gauss-Bonnet theorem, the former of which has analogues even in network optimization and has a nice formulation (and proof) in terms of differential forms.

Some proofs are in:

Rudin's is as usual very clean and readable. Buck's/Arnoux's approach is very approachable and standard. It that is too geometric, try one like Flanders'. Boothby's Intro to Differentiable Manifolds and Riemannian Geometry (for example) has a relatively short proof using differential geometry. In general, the differential forms/geometry approaches are more analytic, but all rely on some way of decomposing regions (or some theorem or definition that treats the same concept) into simpler regions, which is natural when the integral is defined in terms of a Riemann sum.

At the risk of oversimplifying, I present a conceptual overview of a proof. The basic idea of Green's theorem is to see how to generalize the fundamental idea of calculus (FTOC) to several variables. In addition to its assumptions, we add one about the region $\overline{U}\subset\mathbb{R}^n$, e.g. that it is compact, which means closed and bounded in $\mathbb{R}^n$, or a union of bounded and simply connected regions; the exact assumption depends on the method of the proof and its requirements. But all the proofs that I have seen boil down to a decomposition of $U$ into a countable union $\cup_{\alpha}U_\alpha$ of disjoint subregions which are more amenable to applying the FTOC. The subregions are "amenable" in two senses: firstly, they can be sliced in the same direction parallel to a coordinate axis, thereby eliminating one space variable in a portion of $\int_Ud\omega$ to obtain a corresponding portion of $\int_{\partial U}\omega$. But secondly, they are adjacent.

$$ U=\cup_\alpha U_\alpha \qquad U_\alpha \text{ pairwise disjoint, convex} $$ $$ \omega=\sum_{i=1}^{n}\omega_i \qquad \omega_i= u\,dx^1 \wedge\cdots\wedge dx^{i-1} \wedge dx^{i+1} \wedge\cdots\wedge dx^n $$ $$ d\omega=\sum_{i=1}^n d\omega_i%=\nabla\cdotu\, dx^1\wedge\cdots\wedge dx^n \qquad d\omega_i=\frac{\partial u}{\partial x^i}\,dx^1\wedge\cdots\wedge dx^n$$ $$ \eqalign{ \int_{ U } d\omega &= \sum_i \int_{ U } d\omega_i = \sum_i \sum_\alpha \int_{ U_\alpha} d\omega_i \\&= \sum_i \sum_\alpha \int_{\partial U_\alpha} \omega_i = \sum_i \int_{\partial U } \omega_i &= \int_{\partial U } \omega } $$ The integral along the interior boundaries cancel out, due to orientation and adjacency. The above is a conceptual sketch of the derivation of Green's and Stokes' theorems. The portions to the right of a summation over $i$, dealing with the integrals $\int_{R}d\omega_i$ and $\int_{\partial R}\omega_i$ (for $R=U$ or $U_\alpha$), is Green's theorem. Typically, it is then used to derive Stokes' theorem.

  • 1
    $\begingroup$ So what do I make my differential form? If $\omega=udx_{1}\wedge\cdots\wedge dx_{i-1}\wedge dx_{i+1}\wedge\cdots\wedge dx_n$, then $\mathrm{d}\omega = \frac{\partial u}{\partial x_i}dx_1\wedge\cdots\wedge dx_n$. But then how does $\int_{\partial U}\omega = \int_{\partial U}{u\nu^{i}dS}$? $\endgroup$
    – Patch
    Feb 29, 2012 at 16:42
  • 1
    $\begingroup$ This was not a proof, but rather a statement telling me it is a particular instance of a more general theorem. I'd appreciate being shown the connection explicitly... $\endgroup$
    – Patch
    Mar 1, 2012 at 8:07
  • 1
    $\begingroup$ Put an $i$ subscript on your $\omega$ and call $\omega=\sum\omega_i$. The conclusion is Green's theorem: $\int_{\partial U}\omega_i=\int_{U}d\omega_i$. Summing over $i$, we get Stokes' theorem: $\int_{\partial U}\omega=\int_{U}d\omega$. $\endgroup$
    – bgins
    Mar 1, 2012 at 10:50
  • $\begingroup$ Rudin only deals with the 3d case and the sets he deals with seems to be very restrictive. He does not use the "outward normal" language as well. It really bothers me since I have tried for a very long time to locate a resource that deals with the exact version of Evan's calculus theorems in the appendix, but none of them serves that purpose. $\endgroup$
    – koch
    Jun 19, 2018 at 14:33

See Theorem A8.8 on page 270 of Linear Functional Analysis: An Application-Oriented Introduction (Chapt. 8) by Hans Wilhelm Alt for a direct proof. One can skip all the approximation arguments if you assume the boundary is $C^1$ and that $u$ is $C^1$ on a neighborhood of $\overline{U}$. Essentially, you use a partition of unity to reduce the the case where $u$ is supported in a local graph boundary neighborhood, and then it's just a direct computation.


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