Product Rule for Divergence

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Theorem

Let $\map {\mathbf V} {x_1, x_2, \ldots, x_n}$ be a vector space of $n$ dimensions.

Let $\mathbf A$ be a vector field over $\mathbf V$.

Let $U$ be a scalar field over $\mathbf V$.


Then:

$\map {\operatorname {div} } {U \mathbf A} = \map U {\operatorname {div} \mathbf A} + \mathbf A \cdot \grad U$

where

$\operatorname {div}$ denotes the divergence operator
$\grad$ denotes the gradient operator
$\cdot$ denotes dot product.


Proof

From Divergence Operator on Vector Space is Dot Product of Del Operator and definition of the gradient operator:

\(\ds \operatorname {div} \mathbf V\) \(=\) \(\ds \nabla \cdot \mathbf V\)
\(\ds \grad \mathbf U\) \(=\) \(\ds \nabla U\)

where $\nabla$ denotes the del operator.


Hence we are to demonstrate that:

$\nabla \cdot \paren {U \, \mathbf A} = \map U {\nabla \cdot \mathbf A} + \paren {\nabla U} \cdot \mathbf A$


Let $\mathbf A$ be expressed as a vector-valued function on $\mathbf V$:

$\mathbf A := \tuple {\map {A_1} {\mathbf r}, \map {A_2} {\mathbf r}, \ldots, \map {A_n} {\mathbf r} }$

where $\mathbf r = \tuple {x_1, x_2, \ldots, x_n}$ is an arbitrary element of $\mathbf V$.


Let $\tuple {\mathbf e_1, \mathbf e_2, \ldots, \mathbf e_n}$ be the standard ordered basis of $\mathbf V$.


Then:

\(\ds \nabla \cdot \paren {U \mathbf A}\) \(=\) \(\ds \sum_{k \mathop = 1}^n \frac {\map \partial {U A_k} } {\partial x_k}\) Definition of Divergence Operator
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^n \paren {U \frac {\partial A_k} {\partial x_k} + \frac {\partial U} {\partial x_k} A_k}\) Product Rule for Derivatives
\(\ds \) \(=\) \(\ds U \sum_{k \mathop = 1}^n \frac {\partial A_k} {\partial x_k} + \sum_{k \mathop = 1}^n \frac {\partial U} {\partial x_k} A_k\)
\(\ds \) \(=\) \(\ds \map U {\nabla \cdot \mathbf A} + \sum_{k \mathop = 1}^n \frac {\partial U} {\partial x_k} A_k\) Definition of Divergence Operator
\(\ds \) \(=\) \(\ds \map U {\nabla \cdot \mathbf A} + \paren {\sum_{k \mathop = 1}^n \frac {\partial U} {\partial x_k} \mathbf e_k} \cdot \paren {\sum_{k \mathop = 1}^n A_k \mathbf e_k}\) Definition of Dot Product
\(\ds \) \(=\) \(\ds \map U {\nabla \cdot \mathbf A} + \paren {\nabla U} \cdot \mathbf A\) Definition of Gradient Operator, Definition of Vector

$\blacksquare$


Also presented as

This result can also be presented as:

$\nabla \cdot \paren {U \, \mathbf A} = \map U {\nabla \cdot \mathbf A} + \paren {\nabla U} \cdot \mathbf A$

presupposing the implementations of $\operatorname {div}$ and $\grad$ as operations using the del operator.


Sources