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Gauss–Green formula on an affine simplex

Statement

Assume Countable Choice. Let $K$ be a nondegenerate affine $d$ -simplex and let $G \in C^{1} (\overline{K}; R^{d})$ . If $n_{K, F}$ is the outward unit normal on each facet $F$ , then $\int_{K} div G d x = \sum_{F \subset \partial K} \int_{F} G \cdot n_{K, F} d S .$ For an orientation-preserving affine map $F_{K} (\hat{x}) = b + A \hat{x}$ , the oriented area element transforms as $n_{K, F} d S = \det (A) A^{- T} {\hat{n}}_{\hat{F}} d \hat{S} .$

Facts & Assumptions

Given: Countable Choice, a nondegenerate affine simplex $K$ , its fixed reference simplex, and a continuously differentiable vector field on $\overline{K}$ .

[F1]

The reference open simplex is ${{\hat{x}}_{i} > 0, \sum_{i} {\hat{x}}_{i} < 1}$ ; every physical cell is its affine image. (Matching affine simplicial meshes and shape regularity)

[F2]

For affine maps, weak derivatives transform by the constant Jacobian chain rule, and volume measure transforms by $∣ \det A ∣$ . (Affine pullback of weak derivatives and Sobolev seminorms)

[F3]

The one-dimensional fundamental theorem of calculus gives $\int_{a}^{b} f^{'} (t) d t = f (b) - f (a)$ for a continuously differentiable vector-valued function. (If $f : [a, b] \to R^{m}$ is differentiable with integrable $f^{'}$ then $\int_{a}^{b} f^{'} = f (b) - f (a)$ ; and a bounded derivative makes $f$ Lipschitz)

[F4]

Fubini's theorem permits iterated integration of the continuous bounded functions on the bounded simplex used below. (Tonelli and Fubini for the completed product, with only almost-everywhere section measurability)

[F5]

A C1 diffeomorphism obeys Lebesgue change of variables with the absolute determinant. For the bounded integrands here, apply the nonnegative formula separately to positive and negative parts. (A C^1 diffeomorphism satisfies the change-of-variables formula for nonnegative Lebesgue measurable functions)

[F6]

The cited change-of-variables theorem assumes Countable Choice. (The Axiom of Countable Choice ( ${A C}_{ω}$ ))

Proof

1.1F1F3F4givenalgebra

First take ${\hat{S}}_{d}$ . Fix $i$ and write $y = ({\hat{x}}_{j})_{j \neq i}$ , where $y_{j} > 0$ and $\sum_{j \neq i} y_{j} < 1$ . The $i$ th coordinate ranges from $0$ to $g (y) : = 1 - \sum_{j \neq i} y_{j}$ . By [F3], $\int_{0}^{g (y)} \partial_{i} G_{i} (y, t) d t = G_{i} (y, g (y)) - G_{i} (y, 0) .$ Integrating in $y$ and using [F4] gives the volume integral of $\partial_{i} G_{i}$ as the difference of these two graph integrals. The lower graph is the coordinate facet with outward normal $- e_{i}$ and area factor $1$ , hence its flux contribution is $- G_{i} (y, 0)$ . The upper facet is the graph ${\hat{x}}_{i} = 1 - \sum_{j \neq i} {\hat{x}}_{j}$ : its tangent Gram matrix is $I_{d - 1} + 11^{T}$ , whose eigenvalues are $1$ (multiplicity $d - 2$ ) and $d$ , so its area factor is $\sqrt{d}$ . Its outward unit normal is $d^{- 1 / 2} (1, \dots, 1)$ , so the $i$ th component of normal times area factor is $1$ . Thus the upper graph integral is exactly its $i$ th flux contribution. Other coordinate facets have zero $i$ th normal component. Summing over $i$ accounts for every component of the boundary flux and proves the formula on ${\hat{S}}_{d}$ .

1.2F1givenalgebra

Let ${\hat{τ}}_{1}, \dots, {\hat{τ}}_{d - 1}$ be an oriented tangent basis for a reference facet, with orientation chosen so its oriented area vector is $\hat{n} d \hat{S}$ . By the determinant definition of oriented area vector, for every $w \in R^{d}$ , $w \cdot (n_{K, F} d S) = \det (A {\hat{τ}}_{1}, \dots, A {\hat{τ}}_{d - 1}, w) = \det (A) \det ({\hat{τ}}_{1}, \dots, {\hat{τ}}_{d - 1}, A^{- 1} w) = w \cdot (\det (A) A^{- T} \hat{n} d \hat{S}) .$ Equality against every $w$ gives the cofactor area-vector formula. Because $\det A > 0$ , $A^{- T} \hat{n}$ points to the exterior side corresponding to $\hat{n}$ , so the formula has the outward orientation.

2.1F2F5F6step 1.1step 1.2algebra∎

Define $\hat{G} (\hat{x}) : = \det (A) A^{- 1} G (b + A \hat{x})$ . The chain rule gives ${div}_{\hat{x}} \hat{G} = \det (A) \sum_{j, i, l} (A^{- 1})_{j i} (\partial_{x_{l}} G_{i}) (F (\hat{x})) A_{l j} = \det (A) ({div}_{x} G) (F (\hat{x})),$ since $\sum_{j} A_{l j} (A^{- 1})_{j i} = δ_{l i}$ . Applying the reference identity of step 1.1 to $\hat{G}$ , and using [F5] to change variables in its bounded volume integrand, gives the physical volume integral. On each facet, step 1.2 gives $\hat{G} \cdot \hat{n} d \hat{S} = G (F (\hat{x})) \cdot \det (A) A^{- T} \hat{n} d \hat{S} = G \cdot n_{K, F} d S .$ Summing the facet identities proves the stated Gauss–Green formula on $K$ .

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Sources

Jean-Luc Guermond, MATH 661 Fall 2025, Chapter 14 (standard reference, not scraped)
Joachim Schöberl, Numerical Methods for Partial Differential Equations, Chapter 6 (standard reference, not scraped)