172 reputation
5
bio website
location
age
visits member for 2 years, 6 months
seen Oct 11 at 8:15

Mar
14
comment Looking for a function
Thanks, I found $\frac{\tanh(Cx)}{\tanh(C)}$ satisfies my needs.
Jan
14
comment Navier-Stokes Equation and turbulence, current status of research?
Thanks. I have already bookmarked it. I've been following this news for many days. :)
Jun
11
comment Looking for a 3D smooth step function
I am just using the region between -1<x<+1. I should say, it is just step-like function, not a perfect alternative for step function though.
May
29
comment How to apply chain rule to this term $\vec a \cdot \nabla \cdot \nabla \vec b$?
So, if I understand correctly, you do not suggest using notation like "$\left( {\nabla \vec b} \right) \cdot \left( {\nabla \vec a} \right)$", because they are not well defined and is not powerful enough to represent the tensor calculus. Please correct me if I am wrong. But if I do use the notation of $\left( {\nabla \vec b} \right) \cdot \left( {\nabla \vec a} \right)$ to represent $\vec\nabla a_i)(\vec\nabla b_i$, then it is OKAY to write $$\vec a\cdot\nabla\cdot\nabla\vec b=\nabla \cdot \left({\vec a\cdot\nabla\vec b}\right) - \left({\nabla \vec b} \right)\cdot\left({\nabla\vec a}\right)$$?
May
29
comment How to apply chain rule to this term $\vec a \cdot \nabla \cdot \nabla \vec b$?
$\vec a$ 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 ($\left( {\nabla \vec b} \right) \cdot \left( {\nabla \vec a} \right)$)?
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
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
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. :)