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Green's second identity on a glued elementary solid region

Statement

Let a finite gluing of elementary solid regions be given, with union $E$ and outer boundary presentation $Σ^{o u t}$ , let $O$ be an open set containing $E$ , and let $u, v : O \to R$ both be $C^{2}$ . Then

$∭_{E} (u Δ v - v Δ u) = \iint_{\partial E} (u ⟨ \nabla v, n ⟩ - v ⟨ \nabla u, n ⟩) .$

Both functions are required to be $C^{2}$ , which is a stronger hypothesis than the first identity places on either of them.

Facts & Assumptions

Given: The finite gluing with union $E$ and outer presentation $Σ^{o u t}$ , the open $O \supseteq E$ , and the $C^{2}$ functions $u, v$ on $O$ .

[F1]

For scalar-valued $f$ the gradient is $\nabla f = (\partial_{0} f, \dots, \partial_{m - 1} f)$ (The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case), and $Δ f = div \nabla f$ for $C^{2}$ $f$ (The Laplacian of a $C^{2}$ function and of a $C^{2}$ vector field).

[F2]

For $x, y \in R^{m}$ , $⟨ x, y ⟩ = \sum_{i < m} x_{i} y_{i}$ ; in particular $⟨ x, y ⟩ = ⟨ y, x ⟩$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

[F3]

A scalar $f$ is of class $C^{k}$ on $U$ when every iterated derivative of length at most $k$ exists and is continuous on $U$ ; in particular a $C^{2}$ function is $C^{1}$ ( $C^{k}$ maps and multi-index derivative notation in Euclidean space).

[F4]

The flux over a finite patch presentation is a finite sum of parameter integrals of continuous integrands (The divergence theorem for finite gluings of elementary solid regions).

[L1]

Under the hypotheses above with $u$ of class $C^{1}$ and $v$ of class $C^{2}$ , $∭_{E} (⟨ \nabla u, \nabla v ⟩ + u Δ v) = \iint_{\partial E} u ⟨ \nabla v, n ⟩$ (Green's first identity on a glued elementary solid region).

[L2]

For integrable $f, g$ on a nondegenerate rectangle and scalars $α, β$ , the function $α f + β g$ is integrable with integral $α \int f + β \int g$ (Linearity, monotonicity, the absolute-value estimate and coordinate-slice additivity for the Riemann integral in $R^{m}$ ).

[L3]

Every continuous real function on a compact Jordan measurable set is Riemann integrable over it (A continuous real function on a compact Jordan measurable set is Riemann integrable over that set), and for a finite gluing $E$ is compact and Jordan measurable (Internal faces cancel and volume integrals add when elementary solid regions are glued).

Proof

technique · direct

1.1

Both $u$ and $v$ are $C^{2}$ on $O$ , hence also $C^{1}$ there by [F3]. So [L1] applies as it stands and gives $∭_{E} (⟨ \nabla u, \nabla v ⟩ + u Δ v) = \iint_{\partial E} u ⟨ \nabla v, n ⟩$ ; and it applies again with the roles of the two functions exchanged, which is legitimate exactly because both are $C^{2}$ , giving $∭_{E} (⟨ \nabla v, \nabla u ⟩ + v Δ u) = \iint_{\partial E} v ⟨ \nabla u, n ⟩$ .

givenF3L1

2.1

All the integrands appearing in step 1.1 are continuous: $\nabla u$ and $\nabla v$ have continuous components by [F1] and [F3], $Δ u$ and $Δ v$ are continuous by [F1] and [F3], and each boundary integrand is a continuous function on a compact Jordan parameter region by [F4]. So every one of them is integrable over the relevant set by [L3], and differences of them may be taken inside the integrals by [L2].

givenF1F3F4L2L3

3.1

Subtract the second identity of step 1.1 from the first, using step 2.1 to combine the integrals. By the symmetry of the inner product in [F2] the two terms $⟨ \nabla u, \nabla v ⟩$ and $⟨ \nabla v, \nabla u ⟩$ are equal and cancel, leaving $∭_{E} (u Δ v - v Δ u)$ on the left and $\iint_{\partial E} (u ⟨ \nabla v, n ⟩ - v ⟨ \nabla u, n ⟩)$ on the right.

step 1.1step 2.1F2L2∎

Remarks

What the extra hypothesis buys. The first identity needs only one of the two functions to be $C^{2}$ ; using it twice with the roles exchanged needs both. That is the whole difference between the two identities, and it is why the second is stated separately rather than as a rearrangement of the first.
The cancellation is the symmetry of the inner product, nothing more. No integration by parts and no mixed-partials theorem enters here: the term that cancels is literally the same function written two ways.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

J. Feldman, A. Rechnitzer and E. Yeager, CLP-4 Vector Calculus (University of British Columbia), section 4.2 (standard reference, not scraped)
M. Corral, Vector Calculus, chapter 4 (LibreTexts) (standard reference, not scraped)