## Identity for Product of a Function With a Vector Field Satisfying Poisson's Equations

Theorem
If
$f$
is a function and
$\mathbf{v}$
is a vector field satisfying
$\nabla^2 (f \mathbf{v}) =0$
and
$\mathbf{\nabla} \times (\mathbf{\nabla} \times (f \mathbf{v}))=0$
then
$\mathbf{\nabla} ((\mathbf{\nabla} f) \cdot \mathbf{v} + f \mathbf{\nabla} \cdot (\mathbf{v})) =0$

Proof
$\mathbf{\nabla} \times (\mathbf{\nabla} \times (f \mathbf{v}))= \mathbf{\nabla} (\mathbf{\nabla} \cdot (f \mathbf{v}))- \nabla^2 (f \mathbf{v})$

$\nabla^2 (f \mathbf{v}) =\mathbf{\nabla} \times (\mathbf{\nabla} \times (f \mathbf{v}))=0$
&nbsp so
$\mathbf{\nabla} (\mathbf{\nabla} \cdot (f \mathbf{v} ))=0$

Now use the identity
$\mathbf{\nabla} \cdot (f \mathbf{v}) =(\mathbf{\nabla} f ) \cdot \mathbf{v} + f ( \mathbf{\nabla} \cdot \mathbf{v})$
to obtain
$\mathbf{\nabla} ((\mathbf{\nabla} f ) \cdot \mathbf{v} + f \mathbf{\nabla} \cdot \mathbf{v}) =0$
{"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}