Derivation of Gradient in SPH

sckulp · August 30, 2012, 19:49

I'm having a tricky time with the gradient using the SPH method. The standard SPH formulation makes sense:

$A(\mathbf{r})=\int A(\mathbf{r'})W(\mathbf{r}-\mathbf{r'},h)d\mathbf{r'}$

To find the gradient, integrate by parts:

$\nabla A(\mathbf{r})=\int \nabla A(\mathbf{r'})W(\mathbf{r}-\mathbf{r'},h)d\mathbf{r'}$

$=\int\limits_{\Gamma} A(\mathbf{r'})W(\mathbf{r}-\mathbf{r'},h)d\Gamma-\int A(\mathbf{r'})\nabla W(\mathbf{r}-\mathbf{r'},h)d\mathbf{r'}$

The first term is zero, since W(r,h)=0 for r>=h. So, we are left with

$\nabla A(\mathbf{r})=-\int A(\mathbf{r'})\nabla W(\mathbf{r}-\mathbf{r'},h)d\mathbf{r'}\approx -\sum\limits_j \frac{m_j}{\rho_j}A(\mathbf{r}_j)\nabla W(\mathbf{r}-\mathbf{r_j},h)$

However, the papers I am looking at [1] state

$\nabla A(\mathbf{r})\approx \sum\limits_j \frac{m_j}{\rho_j}A(\mathbf{r}_j)\nabla W(\mathbf{r}-\mathbf{r_j},h)$

That is, there is no minus sign. I feel that I'm missing some stupid mistake, but I've been looking at it for a while, and I don't see it.

Thanks in advance!

[1] J.J. Monaghan. "Smoothed Particle Hydrodynamics and Its Diverse Applications." Annual Review of Fluid Mechanics, Vol. 44: 323 -346. January 2012

mehdi kamyabi · August 31, 2012, 08:51

Dear Scott Kulp

It came from a little astute trick, I advise you to read page 43 and 44 of the book "Smoothed particle hydrodynamic particle a meshfree particle method" by G.R.Liu and M.B Liu in order to understand what happened.

sckulp · August 31, 2012, 10:46

This book is really nice and very informative. I see that equation 2.22 is basically the same as I wrote above (though 2.22 is the divergence), with the negative, and I do understand that. However, in the summary on page 44, equation 2.25 is not negative. I don't see any explanation on why the negative was removed... Any thoughts?

Thanks!

mehdi kamyabi · August 31, 2012, 12:20

I understood your question first time,however I think the answer is in page 44 where the authors wrote : "it should be noted that grad(i)Wij is taken with respect to particle i so the negative sign in equation (2.21) is removed in equation (2.25)."

I agree with you there is an vagueness in the notation formulas,maybe we can say the mistake, but the concept is true, W is an even function so grad(W) is an odd one
it means W in the equation 2.25 is the symmetry as W in equation 2.22 because the first is about all neighbors(which detected by j counters) and second is about the determined particle itself (i), maybe if you put Wji instead of Wij in eq 2.22 everything become correct.

August 30, 2012, 19:49	Derivation of Gradient in SPH	#1
sckulp New Member Scott Kulp Join Date: Feb 2012 Posts: 3 Rep Power: 14	I'm having a tricky time with the gradient using the SPH method. The standard SPH formulation makes sense: $A(\mathbf{r})=\int A(\mathbf{r'})W(\mathbf{r}-\mathbf{r'},h)d\mathbf{r'}$ To find the gradient, integrate by parts: $\nabla A(\mathbf{r})=\int \nabla A(\mathbf{r'})W(\mathbf{r}-\mathbf{r'},h)d\mathbf{r'}$ $=\int\limits_{\Gamma} A(\mathbf{r'})W(\mathbf{r}-\mathbf{r'},h)d\Gamma-\int A(\mathbf{r'})\nabla W(\mathbf{r}-\mathbf{r'},h)d\mathbf{r'}$ The first term is zero, since W(r,h)=0 for r>=h. So, we are left with $\nabla A(\mathbf{r})=-\int A(\mathbf{r'})\nabla W(\mathbf{r}-\mathbf{r'},h)d\mathbf{r'}\approx -\sum\limits_j \frac{m_j}{\rho_j}A(\mathbf{r}_j)\nabla W(\mathbf{r}-\mathbf{r_j},h)$ However, the papers I am looking at [1] state $\nabla A(\mathbf{r})\approx \sum\limits_j \frac{m_j}{\rho_j}A(\mathbf{r}_j)\nabla W(\mathbf{r}-\mathbf{r_j},h)$ That is, there is no minus sign. I feel that I'm missing some stupid mistake, but I've been looking at it for a while, and I don't see it. Thanks in advance! [1] J.J. Monaghan. "Smoothed Particle Hydrodynamics and Its Diverse Applications." Annual Review of Fluid Mechanics, Vol. 44: 323 -346. January 2012

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August 31, 2012, 08:51		#2
mehdi kamyabi New Member mehdi Join Date: Nov 2010 Location: Tehran Posts: 16 Rep Power: 16	Dear Scott Kulp It came from a little astute trick, I advise you to read page 43 and 44 of the book "Smoothed particle hydrodynamic particle a meshfree particle method" by G.R.Liu and M.B Liu in order to understand what happened.

August 31, 2012, 10:46		#3
sckulp New Member Scott Kulp Join Date: Feb 2012 Posts: 3 Rep Power: 14	This book is really nice and very informative. I see that equation 2.22 is basically the same as I wrote above (though 2.22 is the divergence), with the negative, and I do understand that. However, in the summary on page 44, equation 2.25 is not negative. I don't see any explanation on why the negative was removed... Any thoughts? Thanks!

August 31, 2012, 12:20		#4
mehdi kamyabi New Member mehdi Join Date: Nov 2010 Location: Tehran Posts: 16 Rep Power: 16	I understood your question first time,however I think the answer is in page 44 where the authors wrote : "it should be noted that grad(i)Wij is taken with respect to particle i so the negative sign in equation (2.21) is removed in equation (2.25)." I agree with you there is an vagueness in the notation formulas,maybe we can say the mistake, but the concept is true, W is an even function so grad(W) is an odd one it means W in the equation 2.25 is the symmetry as W in equation 2.22 because the first is about all neighbors(which detected by j counters) and second is about the determined particle itself (i), maybe if you put Wji instead of Wij in eq 2.22 everything become correct.