$ \nabla \rho \cdot \nabla \Delta \rho=\operatorname{div}\left(\nabla^{2} \rho \nabla \rho\right)-\left|\nabla^{2} \rho\right|^{2} $

Question

In the book Entropy Methods for Diffusive Partial Differential Equations, §2.1, p. 21 it says: $$ \nabla \rho \cdot \nabla \Delta \rho=\operatorname{div}\left(\nabla^{2} \rho \nabla \rho\right)-\left|\nabla^{2} \rho\right|^{2} $$

Can somebody explain where this comes from?

Answer 1

Using the product rule $\nabla \cdot (\phi \vec A)=\nabla \phi\cdot \vec A+\phi \nabla \cdot \vec A$, with $\phi =\nabla^2 \rho$ and $\vec A=\nabla \rho, $we have

$$\begin{align} \nabla \rho \cdot \nabla^2 \nabla \rho& =\underbrace{\nabla \rho}_{\vec A} \cdot \nabla(\underbrace{\nabla^2 \rho}_{=\phi})\\\\ &=\nabla \cdot \left( \underbrace{\nabla \rho}_{\vec A} \,\underbrace{\nabla^2 \rho}_{\phi}\right)-\underbrace{\nabla^2\rho}_{\phi} \,\underbrace{ \nabla^2 \rho}_{\nabla \cdot \vec A}\\\\ &=\nabla \cdot \left( \nabla^2 \rho \nabla \rho\right)-\left|\nabla^2\rho \right|^2 \end{align}$$

as was to be shown!