Physlib.Mathematics.Calculus.Divergence

Let $E$ be a normed space over a field $\mathbf{k}$. The divergence of a map $f: E \to E$ at a point $x \in E$ is defined as the trace of the Fréchet derivative (differential) of $f$ at $x$.

Let $E$ be a normed space over a field $\mathbf{k}$. The divergence of the zero map $f: E \to E$ (defined by $f(x) = 0$ for all $x \in E$) is the zero function, mapping every point in $E$ to $0 \in \mathbf{k}$.

Let $E$ be a normed space over a field $\mathbf{k}$. For any finite set $s$, basis $b = \{b_i\}_{i \in s}$ of $E$, and map $f: E \to E$, the divergence of $f$ at a point $x \in E$ is equal to the sum of the diagonal components of its Fréchet derivative in that basis: $$\text{div} f(x) = \sum_{i \in s} [Df(x)(b_i)]_i$$ where $Df(x)$ is the Fréchet derivative of $f$ at $x$, and $[v]_i$ denotes the $i$-th coordinate of the vector $v$ with respect to the basis $b$.

Let $E$ be a normed space over a field $\mathbf{k}$, and let $b = \{b_i\}_{i \in \iota}$ be a basis of $E$ indexed by a finite type $\iota$. For any map $f : E \to E$, the divergence of $f$ at a point $x \in E$ is equal to the sum of the diagonal components of its Fréchet derivative with respect to the basis $b$: $$\text{div} f(x) = \sum_{i \in \iota} [Df(x)(b_i)]_i$$ where $Df(x)$ is the Fréchet derivative of $f$ at $x$, and $[v]_i$ denotes the $i$-th coordinate of the vector $v$ with respect to the basis $b$.

Let $d$ be a dimension and $\text{Space}_d$ be the corresponding $d$-dimensional real normed space. For any map $f: \text{Space}_d \to \text{Space}_d$ that is differentiable over $\mathbb{R}$, the divergence of $f$ (defined as the trace of its Fréchet derivative) is equal to the coordinate-based divergence of its representation in the canonical basis: $$\text{div} f = \text{Space.div}(\text{repr} \circ f)$$ where $\text{repr}$ is the map that assigns to each vector its coordinates with respect to the basis of $\text{Space}_d$, and $\text{Space.div}$ is the divergence operator acting on coordinate-valued functions.

Let $E$ and $F$ be finite-dimensional normed spaces over a field $\mathbf{k}$. Consider the maps $f: E \times F \to E$ and $g: E \times F \to F$. For any point $(x, y) \in E \times F$, if $f$ and $g$ are differentiable at $(x, y)$, then the divergence of the product map $(f, g): E \times F \to E \times F$ at that point is the sum of the partial divergences: $$\text{div} (f, g)(x, y) = \text{div} (x' \mapsto f(x', y))(x) + \text{div} (y' \mapsto g(x, y'))(y)$$ where the first term on the right is the divergence of $f$ restricted to the first component at $x$, and the second term is the divergence of $g$ restricted to the second component at $y$.

Let $E$ be a normed space over a field $\mathbf{k}$. For any maps $f, g: E \to E$ and any point $x \in E$, if $f$ and $g$ are differentiable at $x$, then the divergence of the sum map $f + g$ at $x$ is equal to the sum of the divergences of $f$ and $g$ at $x$: $$\text{div}(f + g)(x) = \text{div} f(x) + \text{div} g(x)$$

Let $E$ be a normed space over a field $\mathbf{k}$. For any map $f: E \to E$ and any point $x \in E$, the divergence of the map $x \mapsto -f(x)$ at $x$ is equal to the negation of the divergence of $f$ at $x$: $$\text{div}(-f)(x) = -\text{div } f(x)$$

Let $E$ be a normed space over a field $\mathbb{k}$. Let $f, g: E \to E$ be maps that are differentiable at a point $x \in E$. Then the divergence of their difference $(f - g)$ at $x$ is equal to the difference of their individual divergences at $x$: $$\text{div}(f - g)(x) = \text{div} f(x) - \text{div} g(x)$$ where the divergence is defined as the trace of the Fréchet derivative.

Let $E$ be a normed space over a field $\mathbf{k}$. Let $f: E \to E$ be a map and $c \in \mathbf{k}$ be a scalar. If $f$ is differentiable at $x \in E$, then the divergence of the map $x \mapsto c \cdot f(x)$ at $x$ is equal to $c$ times the divergence of $f$ at $x$, i.e., $\operatorname{div}(c \cdot f)(x) = c \cdot \operatorname{div} f(x)$.

Let $E$ be a finite-dimensional generalized inner product space over a field $\mathbb{k}$ (where $\mathbb{k}$ is $\mathbb{R}$ or $\mathbb{C}$). Let $f: E \to \mathbb{k}$ be a scalar field and $g: E \to E$ be a vector field, both of which are differentiable at a point $x \in E$. Then the divergence of the scalar-vector product $x \mapsto f(x) \cdot g(x)$ at $x$ is given by: $$\text{div}(f \cdot g)(x) = f(x) \cdot \text{div} g(x) + \langle \nabla f(x), g(x) \rangle$$ where $\text{div}$ is the divergence (defined as the trace of the Fréchet derivative), $\nabla f(x)$ is the gradient of $f$ at $x$ (represented by the adjoint of the Fréchet derivative of $f$ at $x$ applied to $1$), and $\langle \cdot, \cdot \rangle$ denotes the inner product on $E$.