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

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.

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*}

• What does "EXE" mean/stand for here? +1, Thanks. – Robert Lewis Oct 9 '18 at 1:20
• I mean that EXE (exercise) is a preparatory fact. – HK Lee Oct 9 '18 at 1:26

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

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.

• How we can conclude s∇s=∇(s^2/2) ? – Alex Parker Oct 9 '18 at 1:44
• @AliRahmany: recall that $\partial s^2 / \partial x = ds^2 / ds \partial s / \partial x =2s \partial x / \partial x$ by the chain rule; This is an example of the more general rule $\nabla f(u) = df(u)/du \nabla u$ which itself follows from $df(u) / du \langle X, \nabla u \rangle = df(u) / du X[u] = X[f(u)] = \langle X, \nabla f(u) \rangle.$ – Robert Lewis Oct 9 '18 at 2:00

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.