Index notation for vector calculus proof

Question

I’d like to prove that $\nabla v \cdot \nabla w = \frac{1}{2} \Big(\nabla^2(vw) - v\nabla^2 w -w\nabla^2 v\Big)$. I’ve attempted to use index notation, but I am unsure of how to rely on the chain rule to obtain the result. I am unable to intuitively see where the factor of half comes from as well. I seek your help! Thank you!

Answer 1

Generally it's easier to prove these identities starting from the longer side (the right side in this case). Also, make sure you remember the product rule: $$\begin{align*} \nabla^2(vw) - v\nabla^2 w -w\nabla^2 v &= \partial_i^2(vw) - v\partial_i^2w - w\partial_i^2v \\ &= \partial_i(w\partial_i v + v\partial_i w) - v\partial_i^2w - w\partial_i^2v \\ &= (\partial_iw)(\partial_i v) + w\partial_i^2v + (\partial_iv)(\partial_i w) + v\partial_i^2w - v\partial_i^2w - w\partial_i^2v \\ &= 2(\partial_i v)(\partial_i w) \\ &= 2\nabla v \cdot \nabla w \end{align*}$$

Answer 2

The $\vec\nabla$ operator is such that: $$\vec\nabla(\vec u)= \sum_i \frac{\partial u_i}{\partial x_i}\vec {\sf e}_i$$

So it may be treated as a pseudo vector $$\nabla_i( u_i )= \dfrac{\partial~u_i}{\partial x_i~}$$

Then the chain rule is applied as normal for a derivative operator; $$\begin{split}\nabla_i(v_iw_i) &= v_i\,\nabla_i(w_i)+\nabla_i(v_i)\,w_i \\[2ex] \therefore\quad \vec\nabla (\vec v\cdot\vec w) &= \vec v\cdot\vec\nabla(\vec w)+\vec\nabla(\vec v)\cdot\vec w\end{split}$$

---

So, likewise, start at the index notation for $\vec\nabla^2(\vec v\cdot \vec w)$,

$$\nabla_i^2(v_iw_i) {= \nabla_i(\nabla_i(v_iw_i)) \\ =\nabla_i(v_i\nabla_i(w_i)+\nabla_i(v_i)~w_i) \\ = \nabla_i(v_i\nabla_i(w_i))~+~\nabla_i(\nabla_i(v_i)~w_i)}$$

Now just apply the chain rule again on those terms.