Show that $\nabla\cdot [(\vec u\times \vec v)\times \vec w]=\vec w\cdot [(\vec v\cdot \nabla)\vec u-(\vec u\cdot \nabla)\vec v]$

Question

If $\vec u, \vec v~ \text{and} ~\vec w$ are vectors, show that

$$\nabla\cdot [(\vec u\times \vec v)\times \vec w]=\vec w\cdot [(\vec v\cdot \nabla)\vec u-(\vec u\cdot \nabla)\vec v]$$ if $\operatorname{div}\vec u=0=\operatorname{div} \vec v$, $\operatorname{curl} \vec w=\vec{0}$.

Please suggest which identity should I use to start with.

I'm seeking guidance or hints to begin tackling this problem, as I feel unsure and don't know where to start. I'm hoping someone can provide me with some pointers or direct me towards the right path.

Edit: Could anyone recommend an approach to initiate problem-solving without incorporating tensor-based methods?

Answer 1

first you can use levi civita identity to evaluate the cross product in the left hand side,then it will looks like this : div((U x V)i x W) = div((eijk Uj.Vk) x W) ... (i) then i will turn the index "j" in the "Uj" to be "k" using kronecker delta: Uj = δj^ k Uk. then the equation "(i)" will become : div((eijk.δj^ k.Uk.Vk) x w ) = div(eijk.δj^ k (z x w) )... (ii). for z as some scalar from "<u,v>". and if we continue,i can contract the indicies in the levi civita by kronecker delta,then equation "(ii)" become: div(eijj.(z x w)) = 0 why ? remember that repeating indices in levi civita means zero. then it suited to the right hand side.of course when div(U) = div(V) = 0 the right hand side equal to zero too.

and the proof done :) CMIIW