Rotation and divergence of vector field

4.7 Rotation and divergence of vector field

Recall Green formula on the plain (flux – divergence). Assuming $F = (X, Y)^{T}$ , $Ω \subseteq R^{2}$ is bounded and $\partial Ω$ is piecewisely of $C^{1}$ , then

∫ ∫ F ⋅ ndl = div Fdxdy ∂Ω Ω

Theorem 4.7.1 (Gauss formula in physical form) Assuming $Ω \subseteq R^{3}$ is a bounded closed domain, $\partial Ω$ is a surface piecewisely of $C^{1}$ , $n$ is the unit outer norm vector on $\partial Ω$ . Given

F =: Ω → ℝ3

is of $C^{1}$ , then

∫ ∫ F ⋅ nd σ = div Fdxdydz ∂Ω Ω

where

∂X ∂Y ∂Z ∂(X, Y,Z ) divF = ----+ ----+ ----= tr -----------= ∇ ⋅ F ∂x ∂y ∂z ∂ (x, y,z)

See Figure 4.10 for the geometric meaning of Gauss formula. Another meaning of Gauss formula is: take a small ball

B (P_{0}, ϵ)

P_{0} \in Ω

, then we have

1 ∫ div F(P0 ) = lim ---------- F ⋅ nd σ 𝜖→0 |B (P0,𝜖)| ∂B (P0,𝜖)

Figure 4.10: Geometric meaning of Gauss formula

Assuming the form of second-order derivative corresponding with $F \cdot n d σ$ is $ω = X d y \land d z + Y d z \land d x + Z d x \land d y$ , then the exterior differential is

dω = dX ∧ dy ∧ dz + dY ∧ dz ∧ dx + dZ ∧ dx ∧ dy

Since

∂X ∂X ∂X dX = ----dx + ----dy + ----dz ∂x ∂y ∂z

we have $\begin{aligned} d ω & = \frac{\partial X}{\partial x} d x \land d y \land d z + \frac{\partial Y}{\partial y} d y \land d z \land d x + \frac{\partial Z}{\partial z} d z \land d x \land d y \\ = (\frac{\partial X}{\partial x} + \frac{\partial Y}{\partial y} + \frac{\partial Z}{\partial z}) d x \land d y \land d z = (\frac{\partial X}{\partial x} + \frac{\partial Y}{\partial y} + \frac{\partial Z}{\partial z}) d x d y d z \end{aligned}$

where $d x \land d y \land d z = d x d y d z$ is due to the natural positive direction of the space (outer norm vector). So another form of Gauss formula is

Theorem 4.7.2 (Gauss formula in mathematical form) $\int_{\partial Ω} ω = \int_{Ω} d ω$ .

Many formulas has the form above.

N-L formula (oriented, integral of the second kind). Assuming $Ω = [a, b]$ , then $\partial Ω = {a, b}$ ; the outer norm vector of $\partial Ω$ points along $+ x$ at $x = b$ , along $- x$ at $x = a$ . So
$∫ b ∫ ∫ f(x)dx = dF = F = F (b) ⋅ 1 + F (a) ⋅ (− 1 ) = F (b) − F(a) a Ω ∂Ω$
Assuming $γ \subseteq R^{m}$ is a curve of $C^{1}$ , then
$∫ γ dF = F(B ) − F (A )$
where $A, B$ is the starting point and destination of $γ$ .
Green formula (flux – divergence).
$∮ ∫ ∫ F ⋅ ndl = div Fd σ = ∇ ⋅ Fdxdy ∂D D D$
Written in the form of derivative, we have
$∮ ∫ ∂X-- ∂Y-- X◟dy--−◝◜Y-dx◞= ∖left(∂x + ∂y ∖right)dxdy ∂D ω D ◟------------◝◜------------◞ dω$
Gauss formula. Assuming $Ω \subseteq R^{3}$ is a bounded closed domain, we have
$∮ ∫ ∫ F ⋅ ndσ = div Fd ρ = ∇ ⋅ Fdxdydz ∂Ω Ω Ω$
Written in the form of derivative, we have
$∮ ∫ Xdy ∧ dz + Y dz ∧ dx + Zdx ∧ dy = ∖left(∂X--+ ∂Y--+ ∂Z-∖right )dxdydz ∂Ω ◟---------------◝◜--------------◞ Ω ◟------∂x----∂y--◝◜∂z---------------◞ ω dω$

Recall the circulation – rotation form of Green formula.

∮ ∮ ∫ ∫ ∫ F ⋅ dl = F ⋅ Tdl = rotFd σ = ∇ × F = ∇ × Fdxdy ∂D ∂D D D D

Written in the form of derivative, we have

∮ ∫ ∂Y-- ∂X-- ∂D X◟dx--+◝◜Y-dy◞ = D ∖lef t(∂x − ∂y ∖right)dx ∧ dy ω ◟ -------------◝◜-------------◞ dω

Theorem 4.7.3 (Stokes formula in physical form)

∮ ∫ ∫ F ⋅ dl = rot F ⋅ nd σ = (∇ × F ) ⋅ nd σ ∂Σ Σ Σ

Theorem 4.7.4 (Stokes formula in mathematical form) $\begin{aligned} \oint_{\partial Σ} \underset{ω}{\underset{⏟}{X d x + Y d y + Z d z}} \\ = \int_{Σ} \underset{d ω}{\underset{⏟}{(\frac{\partial Z}{\partial y} - \frac{\partial Y}{\partial z}) d y \land d z + (\frac{\partial X}{\partial z} - \frac{\partial Z}{\partial x}) d z \land d x + (\frac{\partial Y}{\partial x} - \frac{\partial X}{\partial y}) d x \land d y}} \end{aligned}$

Theorem 4.7.5 (General Stokes formula)

∫ ∫ ω = dω ∂Σ Σ