Let $f$ be a $C^2$ function and let $\textbf{F}$ be a smooth vector field on $\mathbb R^3$.

Question

Let $f$ be a $C^2$ function and let $\textbf{F}$ be a smooth vector field on $\mathbb R^3$.

Explain why div$(\nabla f)=0$ is false.

This was a multiple choice question on a past test. Can someone explain why this is false or drop hints? Thank you.

Answer 1

I'll worry about $\mathbf F$ when I know its role in the question. Meanwhile, $\text{div} (\nabla f) = \nabla \cdot \nabla f = \nabla^2 f = 0$ when $f$ is a harmonic function, by definition of harmonic function. Such functions exist; e.g. take the real and imaginary parts of any holomoprhic function $g(z) = u(x, y) + i v(x, y)$ where $z = x + iy \in \Bbb C^2$; this is of course a standard, elementary and universally known fact to anyone who has begun to study complex analysis. These things being said, most functions aren't harmonic (I leave the meaning of "most" here to the intuition; it can be made rigorous), so the statement that $\text{div} (\nabla f) = 0$ is not generally true. It is generally false. It is easy to find examples of $f$ with $\nabla^2f \ne 0$; just pick a $C^2$ $f$ and evaluate $\nabla^2f$; you'll probably find it is nonzero (and again, "probably" comes with the same caveat as did "most"). Anyway, look at ${x^2 + y^2}$. It is not harmonic. There are many like it in this regard. So $\text{div} (\nabla f) = 0$ is not true for arbitrary $C^2$ $f$.

What is $\mathbf F$?

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!