Divergence of Product of Scalar Field with Gradient of Scalar Field

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Theorem

Let $R$ be a region of space.

Let $U$ and $W$ be scalar fields over $R$.


Then:

$\map {\operatorname {div} } {U \grad W} = U \nabla^2 W + \paren {\grad U} \cdot \paren {\grad W}$

where:

$\operatorname {div}$ denotes the divergence operator
$\grad$ denotes the gradient operator
$\nabla^2$ denotes the Laplacian.


Proof

\(\ds \map {\operatorname {div} } {U \mathbf A}\) \(=\) \(\ds \map U {\operatorname {div} \mathbf A} + \mathbf A \cdot \grad U\) Product Rule for Divergence
\(\ds \leadsto \ \ \) \(\ds \map {\operatorname {div} } {U \grad W}\) \(=\) \(\ds \map U {\operatorname {div} \grad W} + \paren {\grad W} \cdot \paren {\grad U}\) substituting $\grad W$ for $\mathbf A$
\(\ds \) \(=\) \(\ds U \nabla^2 W + \paren {\grad U} \cdot \paren {\grad W}\) Laplacian on Scalar Field is Divergence of Gradient

$\blacksquare$


Also presented as

This result can also be presented as:

$\map {\operatorname {div} } {U \grad W} = U \nabla^2 W + \nabla U \cdot \nabla W$

presupposing the implementation of $\grad$ as an operation using the del operator.


Sources

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