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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.


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What about computing $\vec{a}\cdot\Delta\vec{b}$? – Raskolnikov May 29 '12 at 18:50
$\nabla \vec{b}$ is conventionally understood as divergence of $\vec{b}$ – Valentin May 29 '12 at 18:54
Did you mean $\vec a\cdot\Delta \vec b = \Delta \left( {\vec a\vec b} \right) - \vec b\cdot\Delta \vec a$ – Daniel May 29 '12 at 18:55
What kind of derivative are you trying to compute? What do $\vec{a}$ and $\vec{b}$ depend on? Scalar variables like $t$, $x$, and $y$? Also, you are going to need some kind of product rule if that left dot is representing a dot product. – alex.jordan May 29 '12 at 18:56
Why do you want to apply the chain rule? I don't see a composition of functions here. In general, these kinds of derivatives are best manipulated using Einstein notation, but I don't think the expression you have can be simplified much further, apart from replacing $\nabla\cdot\nabla$ with the vector Laplacian $\nabla^2$. – Rahul May 29 '12 at 19:00
up vote 1 down vote accepted

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 $$\int \!d^dx\,\vec a \cdot \Delta \vec b = \int\!d^dx\,\nabla\left(\sum_i a_i \nabla b_i + \sum_i b_i \nabla a_i\right)- \int \!d^dx\,\vec b \cdot \Delta \vec a-2\int\!d^dx \sum_i(\vec\nabla a_i)(\vec\nabla b_i). $$

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So, $\vec a \cdot \nabla \cdot \nabla \vec b \ne \nabla \cdot \left( {\vec a \cdot \nabla \vec b} \right) - \vec b \cdot \nabla \cdot \nabla \vec a$? – Daniel May 29 '12 at 19:41
@Daniel: you are right there... Where should the second derivatives of $\vec a$ come from? – Fabian May 29 '12 at 19:43
Does your equation $$\int {d^d}x{\kern 1pt} \vec a\cdot\Delta \vec b = \underbrace {\int {d^d}x{\kern 1pt} \nabla \left( {\sum\limits_i {{a_i}} \nabla {b_i}} \right)}_{{\rm{surface term using Gauss}}} - \int {d^d}x\sum\limits_i {(\vec \nabla {a_i})(\vec \nabla {b_i})}$$ means that $$\vec a \cdot \nabla \cdot \nabla \vec b = \nabla \cdot \left( {\vec a \cdot \nabla \vec b} \right) - \left( {\nabla \vec b} \right) \cdot \left( {\nabla \vec a} \right) ?$$ But is $\left( {\nabla \vec b} \right) \cdot \left( {\nabla \vec a} \right)$ a scalar? – Daniel May 29 '12 at 19:46
@Daniel: it depends what you mean by $(\nabla \vec b)\cdot(\nabla \vec a)$; such expressions are not well defined and thus it is usually best to write them out explicitely. It you mean $\sum_{i,j} (\partial_j a_i) (\partial_j b_i)$ then it means exactly that. – Fabian May 29 '12 at 19:48
$\vec a$ and $\vec b$ are both velocities, in fluid dynamics, so they are vectors, and the gradient of the vector field is a tensor field, and I am wondering what it would mean by dot product of two tensor field ($\left( {\nabla \vec b} \right) \cdot \left( {\nabla \vec a} \right)$)? – Daniel May 29 '12 at 19:51

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