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Transverse vector-calculus identities for the splitting $R^{3} = R^{2} \oplus R e_{z}$

Statement

Let $e_{z}$ be a unit vector of $R^{3}$ and identify $R^{3} = R^{2} \oplus R e_{z}$ with transverse coordinates $(x_{1}, x_{2})$ ; write $\nabla_{t} h = (\partial_{1} h, \partial_{2} h)$ , $Δ_{t} h = \partial_{1}^{2} h + \partial_{2}^{2} h$ and, for a transverse $C^{1}$ map $u_{t}$ with components $(u_{1}, u_{2})$ , ${div}_{t} u_{t} = \partial_{1} u_{1} + \partial_{2} u_{2}$ and ${curl}_{2} u_{t} = \partial_{1} u_{2} - \partial_{2} u_{1}$ ; regard $h$ , $\nabla_{t} h$ , $u_{t}$ and $e_{z} \times u_{t}$ as $z$ -independent fields on $Ω_{t} \times R$ , with $e_{z}$ the third basis vector. Let $h : Ω_{t} \to R$ be $C^{2}$ and $u_{t} : Ω_{t} \to R^{2}$ be $C^{1}$ . Then, with the divergence, curl, cross product and inner product of Divergence and curl of a $C^{1}$ vector field, The cross product in $R^{3}$ , The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ : (a) $div (e_{z} \times \nabla_{t} h) = 0$ ; $curl (e_{z} \times \nabla_{t} h) = e_{z} Δ_{t} h$ ; $e_{z} \times (e_{z} \times \nabla_{t} h) = - \nabla_{t} h$ ; $\nabla_{t} h \cdot (e_{z} \times \nabla_{t} h) = 0$ ; $e_{z} \cdot (e_{z} \times \nabla_{t} h) = 0$ ; $∣ e_{z} \times \nabla_{t} h ∣ = ∣ \nabla_{t} h ∣$ ; (b) $curl (e_{z} h) = - e_{z} \times \nabla_{t} h$ and $div (e_{z} h) = \partial_{3} h$ (which vanishes for $z$ -independent $h$ ); (c) $curl (u_{t}) = e_{z} {curl}_{2} u_{t}$ , $div (u_{t}) = {div}_{t} u_{t}$ , and $e_{z} \times u_{t}$ is transverse.

Facts & Assumptions

Given: A unit vector $e_{z} \in R^{3}$ , the orthogonal splitting $R^{3} = R^{2} \oplus R e_{z}$ with transverse coordinates $(x_{1}, x_{2})$ and $e_{z} = e_{3}$ the third standard basis vector; an open set $Ω_{t} \subseteq R^{2}$ ; a $C^{2}$ function $h : Ω_{t} \to R$ and a $C^{1}$ map $u_{t} : Ω_{t} \to R^{2}$ with components $(u_{1}, u_{2})$ ; and the fields $h$ , $\nabla_{t} h$ , $u_{t}$ , $e_{z} \times u_{t}$ regarded as functions on $Ω_{t} \times R$ independent of the third coordinate $x_{3}$ .

[F1]

$\partial_{j} f$ is the derivative of $f$ along the $j$ th coordinate line, and a $C^{1}$ map has continuous first partial derivatives (Directional derivatives and partial derivatives of a map $U \subseteq R^{m} \to R^{n}$ ).

[F2]

For a $C^{1}$ field $F = (F_{1}, F_{2}, F_{3})$ on an open subset of $R^{3}$ , $div F = \partial_{1} F_{1} + \partial_{2} F_{2} + \partial_{3} F_{3}$ and $curl F = (\partial_{2} F_{3} - \partial_{3} F_{2}, \partial_{3} F_{1} - \partial_{1} F_{3}, \partial_{1} F_{2} - \partial_{2} F_{1})$ (Divergence and curl of a $C^{1}$ vector field).

[F3]

The cross product on $R^{3}$ is given by its coordinate formula, it is bilinear and alternating, it is orthogonal to both factors, and the Euclidean inner product is the coordinate dot product $⟨ v, w ⟩ = \sum_{j} v_{j} w_{j}$ (The cross product in $R^{3}$ , The cross product is bilinear, alternating, and orthogonal to both factors, The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

[F4]

The Laplacian of a $C^{2}$ scalar field on an open subset of $R^{n}$ is $Δ f = \partial_{1} \partial_{1} f + \dots + \partial_{n} \partial_{n} f$ (The Laplacian of a $C^{2}$ function and of a $C^{2}$ vector field); in particular $Δ_{t} f = \partial_{1}^{2} f + \partial_{2}^{2} f$ for $f$ on $Ω_{t} \subseteq R^{2}$ .

[F5]

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

[F6]

Derivatives of sums, scalar multiples and products of differentiable real functions obey the sum, scalar and product rules (Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ ); in particular a constant factor and $- \partial_{j} h$ may be moved through $\partial_{i}$ .

Proof

technique · direct: write every object in the orthonormal basis $(e_1,e_2,e_z)$, use the coordinate formulas for the cross product, divergence and curl, and evaluate the coordinate-derivative expressions with the mixed-partials theorem

1.1givenF1F2F3

[Component forms] In the orthonormal basis $(e_{1}, e_{2}, e_{z})$ a transverse vector has the form $v = (v_{1}, v_{2}, 0)$ , and the coordinate formula for the cross product [F3] gives $e_{z} \times v = (e_{z} \times v_{1} e_{1}) + (e_{z} \times v_{2} e_{2}) = v_{1} e_{2} - v_{2} e_{1} = (- v_{2}, v_{1}, 0)$ . Hence $e_{z} \times \nabla_{t} h = (- \partial_{2} h, \partial_{1} h, 0)$ , $e_{z} \times u_{t} = (- u_{2}, u_{1}, 0)$ , and $e_{z} h$ is the field $(0, 0, h)$ ; every field displayed in the Statement is independent of $x_{3}$ , so its $\partial_{3}$ -derivative vanishes wherever the derivative is taken.

2.1step 1.1F2F5F6algebra

[Divergence of $e_{z} \times \nabla_{t} h$ ] By [F2] and step 1.1, $div (e_{z} \times \nabla_{t} h) = \partial_{1} (- \partial_{2} h) + \partial_{2} (\partial_{1} h) + \partial_{3} 0 = - \partial_{1} \partial_{2} h + \partial_{2} \partial_{1} h$ , which is $0$ by [F5] because $h$ is $C^{2}$ ; this is the first identity of clause (a).

2.2step 1.1F2F4F6

[Curl of $e_{z} \times \nabla_{t} h$ ] By [F2] and step 1.1, again with the third-coordinate derivative vanishing, $curl (e_{z} \times \nabla_{t} h) = (\partial_{2} 0 - \partial_{3} \partial_{1} h, \partial_{3} (- \partial_{2} h) - \partial_{1} 0, \partial_{1} \partial_{1} h - \partial_{2} (- \partial_{2} h)) = (0, 0, \partial_{1}^{2} h + \partial_{2}^{2} h) = e_{z} Δ_{t} h$ , which is the second identity of clause (a) by [F4].

2.3step 1.1F3algebra

[Triple product, orthogonality and norm] For an arbitrary transverse $v = (v_{1}, v_{2}, 0)$ , step 1.1 gives $e_{z} \times (e_{z} \times v) = e_{z} \times (- v_{2}, v_{1}, 0) = (- v_{1}, - v_{2}, 0) = - v$ ; the coordinate formulas of [F3] give $v \cdot (e_{z} \times v) = v_{1} (- v_{2}) + v_{2} v_{1} + 0 \cdot 0 = 0$ , $e_{z} \cdot (e_{z} \times v) = 0$ as the third component of $e_{z} \times v$ is $0$ , and $∣ e_{z} \times v ∣^{2} = (- v_{2})^{2} + v_{1}^{2} + 0 = v_{1}^{2} + v_{2}^{2} = ∣ v ∣^{2}$ , both sides nonnegative, so $∣ e_{z} \times v ∣ = ∣ v ∣$ . At $v = \nabla_{t} h$ these are the remaining four identities of clause (a); the case $v = 0$ is included since then both sides vanish.

2.4step 1.1F1F2F6

[Clause (b)] By [F2] and step 1.1, $curl (e_{z} h) = (\partial_{2} h - \partial_{3} 0, \partial_{3} 0 - \partial_{1} h, \partial_{1} 0 - \partial_{2} 0) = (\partial_{2} h, - \partial_{1} h, 0) = - e_{z} \times \nabla_{t} h$ , and $div (e_{z} h) = \partial_{1} 0 + \partial_{2} 0 + \partial_{3} h = \partial_{3} h$ , which is $0$ because $h$ is independent of $x_{3}$ .

3.1step 1.1F1F2∎

[Clause (c)] For $u_{t} = (u_{1}, u_{2}, 0)$ with $\partial_{3} u_{j} = 0$ , [F2] gives $curl (u_{t}) = (\partial_{2} 0 - \partial_{3} u_{2}, \partial_{3} u_{1} - \partial_{1} 0, \partial_{1} u_{2} - \partial_{2} u_{1}) = (0, 0, {curl}_{2} u_{t}) = e_{z} {curl}_{2} u_{t}$ and $div (u_{t}) = \partial_{1} u_{1} + \partial_{2} u_{2} = {div}_{t} u_{t}$ ; and $e_{z} \times u_{t} = (- u_{2}, u_{1}, 0)$ has third component $0$ , so it is transverse.

Transverse vector-calculus identities for the splitting $R^{3} = R^{2} \oplus R e_{z}$

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

Transverse vector-calculus identities for the splitting R3=R2⊕RezR3=R2⊕Rez​

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

Transverse vector-calculus identities for the splitting $R^{3} = R^{2} \oplus R e_{z}$