Daniel
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 May 29 comment How to apply chain rule to this term $\vec a \cdot \nabla \cdot \nabla \vec b$? 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? May 29 comment How to apply chain rule to this term $\vec a \cdot \nabla \cdot \nabla \vec b$? So, $\vec a \cdot \nabla \cdot \nabla \vec b \ne \nabla \cdot \left( {\vec a \cdot \nabla \vec b} \right) - \vec b \cdot \nabla \cdot \nabla \vec a$? May 29 comment How to apply chain rule to this term $\vec a \cdot \nabla \cdot \nabla \vec b$? I have edited the original question. Thanks May 29 revised How to apply chain rule to this term $\vec a \cdot \nabla \cdot \nabla \vec b$? Add more details May 29 comment How to apply chain rule to this term $\vec a \cdot \nabla \cdot \nabla \vec b$? Did you mean $\vec a\cdot\Delta \vec b = \Delta \left( {\vec a\vec b} \right) - \vec b\cdot\Delta \vec a$ May 29 asked How to apply chain rule to this term $\vec a \cdot \nabla \cdot \nabla \vec b$? May 29 comment How to integrate twice of this viscous term? And when you wrote, $$\psi^{n}\nabla\left(2\nu D\left(u\right)\right)=\nabla\left(2\nu\psi^{u}D\left(u\right)\right)-2\nu\nabla‌​\psi^{u}D\left(u\right),$$ it should be actually something like $${\psi ^{\bf{u}}} \cdot \nabla \cdot \left( {2\nu D\left( {\bf{u}} \right)} \right) = blahblah$$, Since $\nabla \cdot \left( {2\nu D\left( {\bf{u}} \right)} \right)$ is kind of laplace operator, right? And so the result would be a vector and then ${\psi ^{\bf{u}}} \cdot \nabla \cdot \left( {2\nu D\left( {\bf{u}} \right)} \right)$ would be a scalar. But how about the right side??? May 29 comment How to integrate twice of this viscous term? And in your equation: $$\nabla\psi^{u}\nabla^{T}u=\nabla\left(\nabla^{T}\psi^{u}u\right)-\nabla\left( \nabla^{T}\psi\right)u$$ What would it mean of "$\psi^{u}u$"? Since both $\psi^{\bf{u}}$ and $\bf{u}$ are vectors... May 27 comment How to integrate twice of this viscous term? I am always wondering if there is any rule of $D$ operator, such as chain rule and any other rules that are applicable to $\nabla$ operator, etc.? :( May 27 comment How to integrate twice of this viscous term? I have just added more details to the question. :) May 27 revised How to integrate twice of this viscous term? added 7 characters in body May 27 awarded Editor May 27 revised How to integrate twice of this viscous term? Add more details May 27 awarded Scholar May 27 accepted How to decompose a divergence operator May 27 asked How to integrate twice of this viscous term? May 25 awarded Student May 25 asked How to decompose a divergence operator