Function time its own gradient and vector scalar its divergence

Question

I wanted to make sure of something, I know that for any function $f$ we have $ff'= \frac{1}{2}(f^2)'$. Then do we have the same principle for the gradient and divergence operators ? Lets take $f$ a function and $\vec{g}$ a vector, do we have the equalities below for any coordinates (spherical, cylindrical?...): $$ f \vec{\nabla}f \stackrel{?}{=} \frac{1}{2}\vec{\nabla}f^2$$ $$\vec{g}.(\vec{\nabla}\cdot \vec{g}) \stackrel{?}{=} \frac{1}{2} \vec{\nabla} \cdot \vec{g^2}$$ and if so does someone have a proof of it or where I could find one ? Because for the first one with the gradient I think this is true but I have big doubts for the one with the divergence. Thanks

Answer 1

$ \newcommand\R{\mathbb R} $

Let $\bullet_s : \R\times\R \to \R$ and $\bullet_v : \R^n\times\R^n \to \R^n$ be commutative products on vectors. A general fact is that the derivative of an expression is the sum of the derivatives of its parts. For $\bullet_s$ and a function $f : \R^n \to \R$ this looks like $$ \nabla(f\bullet_s f) = \dot\nabla(\dot f\bullet_s f) + \dot\nabla(f\bullet_s\dot f) = 2\dot\nabla(\dot f\bullet_s f). \tag{1} $$ The notation $\dot\nabla$ means that we are only differentiating $\dot f$, and the undotted $f$ should be thought of as constant while differentiating. An example of a more verbose notation would be $$ \dot\nabla(\dot f\bullet_s f) = \bigl[\nabla_y(f(y)\bullet_s f(x))\bigr]_{y=x}. $$ When $\bullet_s$ is just multiplication, then (1) gives $$ \nabla f^2 = 2(\nabla f)f = 2f\nabla f. $$

Similarly, when considering $\bullet_v$ and $g : \R^n \to \R^n$ we get $$ \nabla\cdot(g\bullet_v g) = \dot\nabla\cdot(\dot g\bullet_v g) + \dot\nabla\cdot(g\bullet_v\dot g) = 2\dot\nabla\cdot(\dot g\bullet_v g). \tag{2} $$ I stress that this requires that $\bullet_v$ is commutative; if it is anti-commutative like the cross product when $n=3$, a similar derivation to (2) shows that $\nabla\cdot(g\times g) = 0$, as it should since $g\times g = 0$.

The nice thing about the $\dot\nabla$ notation is that it can be treated like a generic vector; any identity that hold for all vectors will also hold when that vector replaced with $\dot\nabla$. So we could say that the reason $\dot\nabla(\dot ff) = f\nabla f$ works is because for any vector $v$ and scalars $a, b$ $$ v(ab) = b(va) $$ so replacing $v$ with $\dot\nabla$, $a$ with $\dot f$, and $b$ with $f$ gives the desired result.

We can use this idea to try to get what you want with $\bullet_v$. However, your equation $g(\nabla\cdot g) = \tfrac12\nabla\cdot g^2$ doesn't make sense as it stands, since the LHS is a vector but the RHS is a scalar. What we'll consider instead is trying to make true $$ \nabla\cdot(g\bullet_v g) = 2F(g)(\nabla\cdot g) \tag{3} $$ where is $F : \R^n \to \R$ and $F(g)$ is $F(g(x))$ with the $x$ dependence suppressed. A sufficient condition is for the following identity on vectors $u,v,w$ to hold: $$ u\cdot(v\bullet_v w) = F(w)(u\cdot v). $$ This fully determines $v\bullet_v w$. Choose $u = e_i$ where $e_i$ is the $i^\text{th}$ standard basis element, multiply by $e_i$, and sum over $i$. This gives $$ v\bullet_v w = F(w)\sum_{i=1}^n(e_i\cdot v)e_i = F(w)v. $$ We need this to be commutative, so $$ F(w)v - F(v)w = 0, $$ But this is impossible unless $F(v) = 0$ for all $v$, since we can simply choose $v$ and $w$ to be linearly independent. (In the case $n=1$ we can choose $F(v) = v$, and $\bullet_v$ and the dot product $\cdot$ are just multiplication.) This suggests (though does not prove) that there is no non-trivial product $\bullet_v$ taking vectors to vectors satisfying (3).