B Calculating the dot Product of \nabla and Vector Identity

[Apashanka]

From the vector identity $\nabla •fA=f(\nabla • A)+A•\nabla f$ where f is a scalar and A is a vector.
Now if f is an operator acting on A how does this formula change??
Like $\nabla •[(v•\nabla)v]$ where v is a vector

 

[Charles Link]

The most sure way of getting the answer would be to write out all of the terms in Cartesian coordinates... I will try to work on it a little and see what I get...$\\$ Edit: With about 10 minutes of work on the above, I believe I get $\nabla \cdot [(\vec{v} \cdot \nabla) \vec{v}]=\frac{1}{2} \nabla^2 (\vec{v}\cdot \vec{v})$. $\\$ I'll be happy to show more detail, but basically, I just worked with Cartesian coordinates.

 

[Charles Link]

A check on my above work shows$\nabla \cdot [(\vec{v} \cdot \nabla) \vec{v}]=\frac{1}{2}\nabla^2 (\vec{v} \cdot \vec{v})-\nabla \cdot (\vec{v} \times \nabla \times \vec{v})$. $\\$ I think the second expression on the right side is zero, but I haven't proven it yet. If the expression in post 2 is not correct, then post 3 contains your answer. I do think post 2 is correct. $\\$ Edit: I made an error in post 2. I believe post 3 is correct. $\\$ To see how I got the result of this post, begin with $\nabla (\vec{v} \cdot \vec{v})$ and use $\nabla (\vec{a} \cdot \vec{b})=(\vec{a} \cdot \nabla ) \vec{b}+(\vec{b} \cdot \nabla )\vec{a}+\vec{a} \times \nabla \times \vec{b}+\vec{b} \times \nabla \times \vec{a}$. $\\$ Next take $\nabla \cdot$ on the expression, to give $\nabla^2 (\vec{v} \cdot \vec{v})$.