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

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