Divergence of the cross product of two gradients

Question

Given are the following functions: $$\varphi_j(\vec{x}), j=1,2$$ and the vector field $\vec{w}$ is the defined as following: $$\vec{w} = \vec{\nabla}\varphi_1 \times \vec{\nabla}\varphi_2$$

what can we say about $\vec{\nabla} \cdot \vec{w}$ (Divergence) ?

Answer 1

It's zero. You have $$ \nabla \cdot (F \times G) = \partial_i \varepsilon_{ijk}F_j G_k = \varepsilon_{kij} G_k\partial_i F_j - \varepsilon_{jik} F_j \partial_i G_k = G \cdot (\nabla \times F) - F \cdot (\nabla \times G), $$ and provided that $\varphi_j$ is continuoously twice-differentiable, $\nabla \times (\nabla \varphi_j) = 0$, so both terms are zero.

Answer 2

Using the product rule, we have

$$\begin{align} \nabla \phi\times\nabla \phi&=\nabla \times (\phi\nabla \phi)-\phi \,\underbrace{\nabla\times\nabla \phi}_{=0}\\\\ &=\nabla \times (\phi\nabla \phi) \end{align}$$

Inasmuch as $\nabla \cdot \nabla \times \vec A=0$, we have

$$\begin{align} \nabla \cdot (\nabla \phi\times\nabla \phi)&=\underbrace{\nabla \cdot (\nabla \times (\phi\nabla \phi))}_{=0}\\\\ &=0 \end{align}$$

as was to be shown!