Divergence (Real Vector Fields)#

Definition

TODO

Notation

\[\operatorname{div} f(\boldsymbol{p}) \qquad \nabla \cdot f (\boldsymbol{p})\]

Divergence

Theorem: Divergence and Partial Derivatives

Let \(f: \mathcal{D} \subseteq \mathbb{R}^n \to \mathbb{R}^n\) be a real vector field.

If \(f\) is totally differentiable at \(\boldsymbol{p} \in \operatorname{int} \mathcal{D}\), then its divergence there is given by the partial derivatives of its component functions \(f_1, \dotsc, f_n\) as follows:

\[\operatorname{div} f = \sum_{k = 1}^n \partial_{k} f_k (\boldsymbol{p})\]

Example: \(F(x,y,z) = \begin{bmatrix}xy & y^2 & xz \end{bmatrix}^{\mathsf{T}}\)

Consider the real vector field \(F: \mathbb{R}^3 \to \mathbb{R}^3\) whose coordinate representation w.r.t. Cartesian coordinates is the following:

\[F(x,y,z) = \begin{bmatrix}xy \\ y^2 \\ xz\end{bmatrix}\]

\[\begin{aligned}\operatorname{div} F(x, y, z) & = \partial_x(xy) + \partial_y (y^2) + \partial_z (xz) \\ & = y + 2y + x \\ & = x + 3y\end{aligned}\]

Proof

TODO

The Divergence Theorem in 2D (Gauß's Theorem in 2D)

Let \(\mathcal{D} \subseteq \mathbb{R}^2\) be bounded and Lebesgue-measurable with a boundary which can be parameterized by a piecewise regular simple closed parametric curve \(\gamma\). Let \(f: \mathcal{D} \to \mathbb{R}^2\) be a real vector field.

If \(f\) is continuously differentiable, then the integral of its divergence on \(\mathcal{D}\) is equal to the line integral of the dot product between \(f\) and \(\gamma\)'s outwards-pointing unit normal vector over \(\gamma\):

\[\iint_{\mathcal{D}} \operatorname{div} f \, \mathrm{d}A = \oint_{\gamma} f \cdot \boldsymbol{n} \, \mathrm{d}s\]

Proof

TODO

The Divergence Theorem in 3D (Gauß's Theorem in 3D)

Let \(\mathcal{D} \subseteq \mathbb{R}^3\) be bounded and Lebesgue-measurable with a boundary which can be parameterized by a piecewise regular simple closed parametric surface \(\mathcal{S}\). Let \(f: \mathcal{D} \to \mathbb{R}^3\) be a real vector field.

If \(f\) is continuously differentiable, then the integral of its divergence on \(\mathcal{D}\) is equal to the surface integral of the dot product between \(f\) and \(\mathcal{S}\)'s outwards-pointing unit normal vector over \(\mathcal{S}\):

\[\iiint_{\mathcal{D}} \operatorname{div} f \, \mathrm{d}\mathcal{D} = \iint_{\mathcal{S}} f \cdot \boldsymbol{n} \, \mathrm{d}\mathcal{S}\]

Proof

TODO