Question on derivation of vector identites and using some symbolic manipulations

Question

Let $f,g : \mathbb R^n \to \mathbb R$, then for the gradient we have the product rule $$ \nabla(fg) = (\nabla f) \cdot g + f \cdot (\nabla g). $$ And by $\Delta(f) = \mbox{div}(\nabla(f)) = \nabla \cdot (\nabla f)$ (viewing $\Delta = (\frac{\partial}{\partial x_1}, \ldots, \frac{\partial}{\partial x_n})$ symbolically as a vector) we have \begin{align*} \Delta(fg) & = \nabla \cdot (\nabla(fg)) \\ & = \nabla \cdot ( \nabla(f) \cdot g + f\cdot \nabla(g) ) \qquad\qquad (*) \\ & = g\cdot ( \nabla \cdot \nabla(f) ) + f \cdot ( \nabla \cdot \nabla(g) ) \qquad (**) \\ & = g\cdot ( \Delta f) + f \cdot ( \Delta g ) \end{align*} by the linearity of the real scalar product. But I doubt this result, because if $n = 1$ this would read as $$ (fg)'' = g\cdot f'' + f\cdot g'' $$ which is wrong, because \begin{align*} (fg)'' & = ((fg)')' \\ & = (fg' + f'g)' \\ & = (fg')' + (f'g)' \\ & = fg'' + f'g' + f''g + f'g' \\ & = fg'' + f''g + 2f'g' \end{align*} So something went wrong with the above derivation, but I cannot see what? Of course something went wrong with the derivation from (*) to (**), but too me it seems valid if I interpret $\Delta \cdot ( ... )$ as a scalarproduct, because $\nabla(f) \cdot g$ is a vector multiplied by a scalar, and for scalar product we have (in the real case) $$ u \cdot (\alpha v + \beta w) = \alpha (u\cdot v) + \beta (u \cdot w) $$ which I simply applied here?

Answer 1

As you suspect, the error is in rewriting $\nabla\cdot(\nabla f \cdot g)$ to $g\cdot (\nabla\cdot \nabla f)$. That would be true if $g$ were a scalar constant, but it is a scalar field, so multiplying by it does not commute with $\nabla$.

For convenience, let's write $F=\nabla f$. Then, using the Leibniz rule we get $$\nabla\cdot(gF) = g(\nabla\cdot F) + (\nabla g) \cdot F $$ (for full rigor we would write this out in coordinates to verify it).

If $g$ were constant, the second term would vanish, but it isn't.

Answer 2

$$\begin{align*} \Delta(fg) & = \nabla \cdot (\nabla(fg)) \\ & = \nabla \cdot ( \nabla(f) \cdot g + f\cdot \nabla(g) ) \\ & = (\nabla \cdot\nabla f)\cdot g+\nabla f\cdot \nabla g+ \nabla f\cdot \nabla g+f\cdot (\nabla \cdot\nabla g) \\ & = g\cdot ( \Delta f) + f \cdot ( \Delta g )+2\nabla f\cdot \nabla g \end{align*}$$