Showing that $∇·(s∇×(s\vec V))=(∇×\vec V)·∇(s^2/2)$

Question

I was given to prove $$∇·(s∇×(s\vec V))=(∇×\vec V)·∇(s^2/2)$$ by using index notations i.e. Kronecker delta and Levi-Civita symbol. But I cannot figure out how I have to tackle the problem to get that $s^2/2$ is right hand side of the equation.

Answer 1

EXE : If $V$ is vector field on $\mathbb{R}^3$, then recall

$$\nabla \cdot (sV)=\nabla s\cdot V + s\nabla\cdot V$$

EXE : $ \nabla \times (sV)= \nabla s\times V +s\nabla\times V $

EXE : $\nabla \cdot (V\times W)=( \nabla \times V)\cdot W - V\cdot (\nabla\times W) $

EXE : $\nabla\cdot (\nabla\times V)=0$

EXE : \begin{align*}&\ \ \nabla\cdot (s\nabla\times (sV)) \\&= \nabla s\cdot (\nabla\times (sV)) +s\nabla\cdot ( \nabla\times (sV)) \\&= \nabla s\cdot (\nabla s\times V +s\nabla \times V) +s\nabla\cdot ( \nabla s\times V +s\nabla \times V ) \\&= \nabla s\cdot (s\nabla \times V) + s \{ ( \nabla\times \nabla s )\cdot V - (\nabla\times V)\cdot \nabla s \} +s\{ \nabla s\cdot (\nabla \times V) +s \nabla\cdot(\nabla\times V) \} \\&= s V\cdot (\nabla\times \nabla s) + s\nabla s\cdot (\nabla \times V)\end{align*}

Answer 2

$\nabla \cdot(s \nabla \times (s\vec V)) \overset{?}{=}(\nabla \times\vec V) \cdot \nabla (s^2/2) \tag 1$

I'm afraid I'm not sufficiently adept with Levi-Civita symbols etc., to easily offer much by way of a derivation using them; I can, however, like my colleague HK Lee, validate the result by means of classic vector analysis:

$s \nabla \times (s \vec V) = s(\nabla s \times \vec V + s\nabla \times \vec V) = s\nabla s \times \vec V + s^2 \nabla \times \vec V; \tag 2$

$\nabla \cdot ( s \nabla \times (s \vec V)) = \nabla \cdot (s\nabla s \times \vec V + s^2\nabla \times \vec V)$ $= \nabla \cdot (s \nabla s \times \vec V) + \nabla \cdot (s^2\nabla \times \vec V); \tag 3$

we make use of the following identity from https://en.m.wikipedia.org/wiki/Vector_calculus_identities:

$\nabla \cdot (A \times B) = B \cdot (\nabla \times A) - A \cdot (\nabla \times B); \tag 4$

thus,

$\nabla \cdot (s \nabla s \times \vec V) = \vec V \cdot (\nabla \times s\nabla s) - s\nabla s \cdot (\nabla \times \vec V); \tag 5$.

now,

$\nabla \times s\nabla s = \nabla \times \nabla \dfrac{s^2}{2} = 0, \tag 6$

since the curl of a gradient always vanishes, identically; thus,

$\nabla \cdot (s \nabla s \times \vec V) = -s\nabla s \cdot (\nabla \times \vec V) = -\nabla(\dfrac{s^2}{2}) \cdot \nabla \times \vec V; \tag 7$

this leaves us with

$\nabla \cdot (s^2 \nabla \times \vec V) = \nabla s^2 \cdot \nabla \times \vec V + s^2 \nabla \cdot (\nabla \times \vec V) = \nabla s^2 \cdot \nabla \times \vec V, \tag 8$

since the divergence of a curl vanishes, always.

Combining (7) and (8) together into (3) we conclude

$\nabla \cdot ( s \nabla \times (s \vec V)) =-\nabla(\dfrac{s^2}{2}) \cdot \nabla \times \vec V + \nabla s^2 \cdot \nabla \times \vec V = \nabla(\dfrac{s^2}{2}) \cdot (\nabla \times \vec V), \tag 9$

as per request.

Answer 3

Summing over repeated indices, the left-hand side is $$\epsilon_{ijk}\partial_i (s\partial_j (sV_k))=\epsilon_{ijk}(\partial_i s\partial_j sV_k+s\partial_i\partial_j sV_k+\partial_i(s^2) \partial_j V_k+s^2 \partial_i\partial_j V_k).$$In the above calculation, which you can follow as an exercise, I've pulled the Levi-Civita symbol outside the derivative and repeatedly used the product rule. Symmetry arguments drop most terms, and we're left with $\partial_i(s^2)\epsilon_{ijk}\partial_j V_k$ as required.