How can div (u cross v) be proved using the product rule?

[oxxiissiixxo]

Please show me the way the proof this!

div (u cross v) = v dot grad (u)- u dot grad(v) where v and u is a vector.

The product rule doesn't seem working

 

[Hootenanny]

Please show me the way the proof this!

div (u cross v) = v dot grad (u)- u dot grad(v) where v and u is a vector.

The product rule doesn't seem working

The "product rule" in vector calculus is different to the product rule for scalar functions. The "vector product rule" depends on type of product you are dealing with (e.g. cross, scalar) and the type of operator that you are dealing with (e.g. divergence, curl). in fact, the relationship that you wish to prove is the product rule for the divergence of a cross product.

The most straightforward way to prove your relation is a by direct computation. You should know that

\boldsymbol{u}\times\boldsymbol{v} = \varepsilon_{ijk}u_jb_k

And that

\text{div}\left(\boldsymbol{A}\right) = \frac{\partial A_i}{\partial x_i}

So all you need to do, is put the two together and compute. (Be careful with your summation indices).

 

[lurflurf]

that should be

\mathbf{\nabla\cdot (u\times v)=v\cdot \nabla\times u-u\cdot \nabla\times v}

div (u cross v) = v dot curl (u)- u dot curl(v)
use partial differentiation
\mathbf{\nabla\cdot (u\times v)=\nabla_u\cdot (u\times v)+\nabla_v \cdot (u\times v)}
div (u cross v)=div_u (u cross v)+div_v (u cross v)
recall
a dot b cross c=-b dot a cross c==c dot a cross b
thus
\mathbf{\nabla_u\cdot (u\times v)=v\cdot \nabla\times u}
\mathbf{\nabla_v\cdot (u\times v)=-u\cdot \nabla\times v}
div_u (u cross v)= v dot curl u
div_v (u cross v)=-u dot curl v

therefore

div (u cross v)=v dot curl (u)- u dot curl(v)

The most straightforward way to prove your relation is a by direct computation.

Introducing an arbitrary coordinate system and a double sum is not very straight forward.
Calculus students hate epsilons and deltas. :)

vector calculus favorite
\mathbf{(a\times\nabla)\times b+a\nabla\cdot b=a\times(\nabla\times b)+(a\cdot\nabla)b}