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 \[\mathbf{\nabla} ((\mathbf{\nabla} f ) \cdot \mathbf{v} + f \mathbf{\nabla} \cdot \mathbf{v}) =0\]