How to apply chain rule to this term $\vec a \cdot \nabla \cdot \nabla \vec b$? Anyone knows how to apply chain rule to this term $\vec a \cdot \nabla  \cdot \nabla \vec b$?
$\nabla$ operator is defined in  Cartesian coordinate system $R^3$ with coordinates $(x, y, z)$, see reference.$\nabla \vec b$ is the gradient of a vector, that is "a tensor", so the divergence of a 2nd order tensor $\nabla  \cdot \nabla \vec b$ is a vector again, and the final dot product of two vectors $\vec a \cdot \nabla  \cdot \nabla \vec b$ would be a scalar.
I'd like to see something similar to $a\nabla b = \nabla (ab) - b\nabla a$, I need this for the purpose of an integration.
Thanks
 A: I guess what you are looking for is a product rule (used for partial integration). For such problems, it is usually very helpful to write the expression explicitly (in coordinates). We have
$$\vec a \cdot \Delta \vec b= \vec a \cdot (\vec\nabla  \cdot \vec\nabla) \vec b = \sum_{i,j}a_i \partial^2_j b_i.$$
Now take a look at
$$
\sum_{i,j}\partial_j (a_i \partial_j b_i)=
\sum_{i,j} \left( a_i \partial^2_j b_i +(\partial_j a_i) (\partial_j b_i)  \right).$$
The formula for partial integration thus reads
$$\int \!d^dx\,\vec a \cdot \Delta \vec b
= \underbrace{\int\!d^dx\,\nabla\left(\sum_i a_i \nabla b_i\right)}_\text{surface term using Gauss} -\int\!d^dx \sum_i(\vec\nabla a_i)(\vec\nabla b_i). $$
Some more (potentially) useful formulas:
Interchanging $a$ and $b$, we have
$$\int \!d^dx\,\vec b \cdot \Delta \vec a
= \underbrace{\int\!d^dx\,\nabla\left(\sum_i b_i \nabla a_i\right)}_\text{surface term using Gauss} -\int\!d^dx \sum_i(\vec\nabla a_i)(\vec\nabla b_i). $$
Subtracting the two relations yields (this is the vector version of Green's second identity) $$\int \!d^dx\,(\vec a \cdot \Delta \vec b- \vec b \cdot \Delta \vec a)
=\int\!d^dx\,\nabla\left(\sum_i a_i \nabla b_i - \sum_i b_i \nabla a_i\right).$$
Adding the two relations yields
$$\int \!d^dx\,\vec a \cdot \Delta \vec b = \int\!d^dx\,\nabla\left(\sum_i a_i \nabla b_i + \sum_i b_i \nabla a_i\right)- \int \!d^dx\,\vec b \cdot \Delta \vec a-2\int\!d^dx \sum_i(\vec\nabla a_i)(\vec\nabla b_i).  $$
