CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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The curl of a curl is the gradient of the divergence minus the Laplacian

Statement

Let $U \subseteq R^{3}$ be open and let $F : U \to R^{3}$ be $C^{2}$ . Then $curl curl F$ , $\nabla div F$ and $Δ F$ are all defined on $U$ and

$curl curl F = \nabla div F - Δ F .$

Here $Δ F$ is the componentwise Laplacian of The Laplacian of a $C^{2}$ function and of a $C^{2}$ vector field.

Facts & Assumptions

Given: The open set $U \subseteq R^{3}$ and the $C^{2}$ field $F : U \to R^{3}$ of the Statement, with the three coordinates named $x, y, z$ .

[F1]

The curl of a $C^{1}$ field $F$ on an open $U \subseteq R^{3}$ is $curl F = (\partial_{y} F_{z} - \partial_{z} F_{y}, \partial_{z} F_{x} - \partial_{x} F_{z}, \partial_{x} F_{y} - \partial_{y} F_{x})$ (Divergence and curl of a $C^{1}$ vector field).

[F2]

The divergence of a $C^{1}$ field $G$ on an open $U \subseteq R^{n}$ is $div G = \sum_{i < n} \partial_{i} G_{i}$ (Divergence and curl of a $C^{1}$ vector field).

[F3]

For a $C^{2}$ map $F$ , $Δ F$ is the field whose $i$ th coordinate is $Δ F_{i}$ , and $Δ f = \sum_{i < n} \partial_{i} \partial_{i} f$ for a $C^{2}$ scalar $f$ (The Laplacian of a $C^{2}$ function and of a $C^{2}$ vector field).

[F4]

For scalar-valued $f$ , its 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).

[F5]

A scalar $f$ is of class $C^{k}$ on $U$ when, for every word $(i_{1}, \dots, i_{r})$ of coordinate indices with $0 \leq r \leq k$ , the iterated derivative $\partial_{i_{r}} \dots \partial_{i_{1}} f$ exists and is continuous on $U$ ( $C^{k}$ maps and multi-index derivative notation in Euclidean space).

[L1]

If $f$ is $C^{2}$ on an open subset of $R^{m}$ , then $\partial_{i} \partial_{j} f = \partial_{j} \partial_{i} f$ for every pair of coordinate indices (Clairaut--Schwarz theorem for continuous second partial derivatives).

Proof

technique · direct

1.1

Every component of $F$ is $C^{2}$ , so by [F5] every iterated derivative $\partial_{i} \partial_{j} F_{a}$ exists and is continuous on $U$ . Hence each coordinate of $curl F$ , being a difference of first partial derivatives of components of $F$ by [F1], has continuous first partial derivatives, so $curl F$ is $C^{1}$ and $curl curl F$ is defined; likewise $div F$ is $C^{1}$ by [F2], so $\nabla div F$ is defined by [F4]; and $Δ F$ is defined by [F3].

givenF1F2F3F4F5

2.1

By [F1] applied twice, the first coordinate of $curl curl F$ is $\partial_{y} (curl F)_{z} - \partial_{z} (curl F)_{y} = \partial_{y} (\partial_{x} F_{y} - \partial_{y} F_{x}) - \partial_{z} (\partial_{z} F_{x} - \partial_{x} F_{z})$ , that is $\partial_{y} \partial_{x} F_{y} - \partial_{y} \partial_{y} F_{x} - \partial_{z} \partial_{z} F_{x} + \partial_{z} \partial_{x} F_{z}$ .

step 1.1F1algebra

3.1

Adding and subtracting the single term $\partial_{x} \partial_{x} F_{x}$ rewrites step 2.1 as $(\partial_{x} \partial_{x} F_{x} + \partial_{y} \partial_{x} F_{y} + \partial_{z} \partial_{x} F_{z}) - (\partial_{x} \partial_{x} F_{x} + \partial_{y} \partial_{y} F_{x} + \partial_{z} \partial_{z} F_{x})$ .

step 2.1algebra

4.1

By [L1], $\partial_{y} \partial_{x} F_{y} = \partial_{x} \partial_{y} F_{y}$ and $\partial_{z} \partial_{x} F_{z} = \partial_{x} \partial_{z} F_{z}$ , so the first bracket of step 3.1 is $\partial_{x} (\partial_{x} F_{x} + \partial_{y} F_{y} + \partial_{z} F_{z}) = \partial_{x} div F$ , the first coordinate of $\nabla div F$ by [F2] and [F4]; the second bracket is $Δ F_{x}$ , the first coordinate of $Δ F$ by [F3]. Hence the first coordinate of $curl curl F$ is that of $\nabla div F - Δ F$ .

step 3.1L1F2F3F4

4.2

In the second coordinate, [F1] gives $\partial_{z} (curl F)_{x} - \partial_{x} (curl F)_{z} = \partial_{z} \partial_{y} F_{z} - \partial_{z} \partial_{z} F_{y} - \partial_{x} \partial_{x} F_{y} + \partial_{x} \partial_{y} F_{x}$ ; adding and subtracting $\partial_{y} \partial_{y} F_{y}$ and applying [L1] to $\partial_{z} \partial_{y} F_{z} = \partial_{y} \partial_{z} F_{z}$ and $\partial_{x} \partial_{y} F_{x} = \partial_{y} \partial_{x} F_{x}$ turns it into $\partial_{y} div F - Δ F_{y}$ . In the third coordinate, [F1] gives $\partial_{x} (curl F)_{y} - \partial_{y} (curl F)_{x} = \partial_{x} \partial_{z} F_{x} - \partial_{x} \partial_{x} F_{z} - \partial_{y} \partial_{y} F_{z} + \partial_{y} \partial_{z} F_{y}$ ; adding and subtracting $\partial_{z} \partial_{z} F_{z}$ and applying [L1] to $\partial_{x} \partial_{z} F_{x} = \partial_{z} \partial_{x} F_{x}$ and $\partial_{y} \partial_{z} F_{y} = \partial_{z} \partial_{y} F_{y}$ turns it into $\partial_{z} div F - Δ F_{z}$ .

step 2.1step 3.1L1F1F2F3F4

5.1

All three coordinates of $curl curl F$ agree with those of $\nabla div F - Δ F$ at every point of $U$ , which is the asserted identity. The hypothesis that $F$ is $C^{2}$ is used in step 1.1, so that all three expressions are defined, and in steps 4.1 and 4.2 as the hypothesis of [L1].

step 4.1step 4.2∎

Remarks

The added and subtracted term is what makes the identity close. The expansion of $(curl curl F)_{x}$ contains no pure second derivative $\partial_{x} \partial_{x} F_{x}$ , while both $\nabla div F$ and $Δ F$ do; that one term belongs to both groups and cancels between them, which is why it can be inserted at will and why neither side alone matches the expansion.

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), Theorem 4.1.7 (standard reference, not scraped)
M. Corral, Vector Calculus, chapter 4 (LibreTexts) (standard reference, not scraped)