network profile
| bio | website | |
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| location | ||
| age | ||
| visits | member for | 1 year, 1 month |
| seen | Jun 12 at 0:09 | |
| stats | profile views | 13 |
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Jun 11 |
accepted | Looking for a 3D smooth step function |
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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. |
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Jun 11 |
revised |
Looking for a 3D smooth step function added 78 characters in body |
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Jun 11 |
asked | Looking for a 3D smooth step function |
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Jun 10 |
awarded | Supporter |
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May 30 |
asked | Navier-Stokes Equation and turbulence, current status of research? |
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May 29 |
accepted | How to apply chain rule to this term $\vec a \cdot \nabla \cdot \nabla \vec b$? |
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May 29 |
awarded | Commentator |
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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 $$\vec a\cdot\nabla\cdot\nabla\vec b=\nabla \cdot \left({\vec a\cdot\nabla\vec b}\right) - \left({\nabla \vec b} \right)\cdot\left({\nabla\vec a}\right)$$? |
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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 ($\left( {\nabla \vec b} \right) \cdot \left( {\nabla \vec a} \right)$)? |
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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? |
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May 29 |
comment |
How to apply chain rule to this term $\vec a \cdot \nabla \cdot \nabla \vec b$? So, $\vec a \cdot \nabla \cdot \nabla \vec b \ne \nabla \cdot \left( {\vec a \cdot \nabla \vec b} \right) - \vec b \cdot \nabla \cdot \nabla \vec a$? |
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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 |
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May 29 |
revised |
How to apply chain rule to this term $\vec a \cdot \nabla \cdot \nabla \vec b$? Add more details |
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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 $\vec a\cdot\Delta \vec b = \Delta \left( {\vec a\vec b} \right) - \vec b\cdot\Delta \vec a$ |
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May 29 |
asked | 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 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??? |
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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... |
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May 27 |
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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.? :( |
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May 27 |
comment |
How to integrate twice of this viscous term? I have just added more details to the question. :) |
