$\nabla\cdot(\phi\nabla\psi) = \phi{\nabla}^2\psi + \nabla\phi\cdot\nabla\psi$: where's the symmetry?

Question

$\nabla\cdot(\phi\nabla\psi) = \phi{\nabla}^2\psi + \nabla\phi\cdot\nabla\psi$ is listed as an identity in Vector_calculus_identities

Trivially rearranging this: $$ \nabla\phi\cdot\nabla\psi = \nabla\cdot(\phi\nabla\psi) - \phi{\nabla}^2\psi$$

Can the RHS be rearranged to show more clearly the expected symmetry between $\phi$ and $\psi$, given they're both scalar functions ?

Answer 1

Either the first two rows of the following \begin{align*} \nabla\phi\cdot\nabla\psi & = \nabla\cdot(\phi\nabla\psi) - \phi{\nabla}^2\psi\\& =\nabla\cdot(\psi\nabla\phi) - \psi{\nabla}^2\phi\\ & =\nabla\cdot(\phi\nabla\psi +\psi\nabla\phi)-\nabla\psi\cdot\nabla\phi-(\phi{\nabla}^2\psi+\phi{\nabla}^2\psi). \end{align*} or using the third: \begin{align*} \nabla\phi\cdot\nabla\psi & =\frac12\left( \nabla\cdot(\phi\nabla\psi +\psi\nabla\phi)-(\phi{\nabla}^2\psi+\phi{\nabla}^2\psi)\right)\\ & = \frac12\left( \nabla^2(\phi\psi)-(\phi{\nabla}^2\psi+\phi{\nabla}^2\psi)\right).\end{align*} Thank you Rahul for pointing out this last equality.

Answer 2

How about something that directly follows:

$$\nabla \cdot (\phi \nabla \psi - \psi \nabla \phi) = \phi \nabla^2 \psi - \psi \nabla^2 \phi $$