Daniel
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 May29 comment How to apply chain rule to this term $\vec a \cdot \nabla \cdot \nabla \vec b$? Did you mean $\vec a\cdot\Delta \vec b = \Delta \left( {\vec a\vec b} \right) - \vec b\cdot\Delta \vec a$ May29 asked How to apply chain rule to this term $\vec a \cdot \nabla \cdot \nabla \vec b$? May29 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??? May29 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... May27 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.? :( May27 comment How to integrate twice of this viscous term? I have just added more details to the question. :) May27 revised How to integrate twice of this viscous term? added 7 characters in body May27 awarded Editor May27 revised How to integrate twice of this viscous term? Add more details May27 awarded Scholar May27 accepted How to decompose a divergence operator May27 asked How to integrate twice of this viscous term? May25 awarded Student May25 asked How to decompose a divergence operator