Is it correct to think of the Laplacian as the divergence of a gradient field?

Question

Factoring out the notation, I see that

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

which looks something like the divergence of the gradient of phi.

Is it actually true?

Thanks,

Answer 1

Yes, it is correct. Indeed, let $\phi \colon \mathbb{R}^N \to \mathbb{R}$ be of class $C^2$, then \begin{align} \operatorname{div}(\nabla\phi) := \sum_{i = 1}^N\frac{\partial}{\partial x_i}(\nabla\phi)_i = \sum_{i = 1}^N\frac{\partial^2 \phi}{\partial x_i^2} =: \Delta\phi. \end{align}