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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 $\vec a \cdot \nabla \cdot \nabla \vec b$?
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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?
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May
27
awarded  Editor
May
27
revised How to integrate twice of this viscous term?
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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