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May 22, 2012, 03:41 

#21 
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Filippo Maria Denaro
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be careful ... for Neumann conditions, the matrix is singular but you have to fulfill the compatibility relation (that involves BCs) such that the system admits infinite solutions...this is an issue very different from the accuracy in solving the system ...


May 22, 2012, 14:44 

#22 
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Yeah the nuemann condition part is can be tricky. I think a lot of people set one of the boundary values to zero instead of using a nuemann condition. You mentioned that you solved this problem in rectangular coordinates. Did you need any special treatment for the neumann condition then?


May 23, 2012, 11:39 

#23  
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Peter
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Quote:
Quote:
In cylindrical coordinates, I have made the Laplacian matrix, except in contrast to the cartesian version it isn't symmetric but is still near singular. I have done the same thing as before and replaced pressure outside the boundary with the first interior point for all the boundary finite differences. Then I try solving the system Ax = b where A is the near singular discrete laplacian matrix, and b is the divergence of the intermediate velocity field. I use the same GMRES method that I used in the cartesian case but I still produce a field that has a divergence on the order of 1E2 to 1E3 when it should be much lower. 

May 23, 2012, 12:01 

#24 
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Well first off if you are solving a poisson type problem with nueman conditions it is 100% singular not just nearly. Try solving a 1D poisson problem with nuemann conditions using gaussian elimination and you'll see what i mean.
Suggestions 1. Examine the residual r = Axb. See if it is maximum where your divergence errors are max. 2. Look at the pressure field itself to see if you really are achieving a nuemann condition on the pressure. 3. Post your MATLAB code, meshplots of pressure, divergence etc. 

May 23, 2012, 13:34 

#25  
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Peter
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Quote:
2. I don't ever calculate the pressure points on the boundary or outside because I am computing on a staggered grid. However based on my formulation of the finite difference matrix I believe that this is enforced. 3. My code can be found in the .zip folder in the link below. If you run the script cartesian_sliding_cavity.m, it will compute the cartesian equivalent to my problem. If you run axisymmetric_sliding_cylinder.m, it will run the code I am currently trying to debug. If anything is unclear, please ask since it will force me to justify my actions and perhaps help me better understand what I am doing wrong. Thanks for all the help https://www.dropbox.com/s/1tiuq4aobu...ric%20Code.zip 

May 23, 2012, 13:52 

#26 
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Filippo Maria Denaro
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do you modify congruently near the boundaries the source term in the Poisson equation?


May 23, 2012, 14:20 

#27 
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Program is surprisingly short, I guess all the built in MATLAB functions help but its also makes it very tough to read/understand.
Anyways % pressure correction rhs = reshape(diff([uW;U;uE])/hx+diff([vS' V vN']')'/hy+avg([vS' V vN']')'./rvp,[],1); [p,flag(k)] = gmres(Lp,rhs,[],[],length(Lp)); . . . div = div_cyl(nx,ny,hx,hy,Ue,Ve,X); maxdiv(k) = max(max(abs(div(2:end1,2:end1)))); It looks like you are using two different methods of calculating the divergence. One way is using diff and that is fed into the poisson solver then you use div_cyl to check the divergence. If you check the diveregence after with the diff method it apperas to be ok. 

May 23, 2012, 16:13 

#28  
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Peter
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Quote:
You are right. I didn't consider that. I guess what confuses me right now is that I expect that the area under the velocity curve should be constant in time. However as time progresses the area under my velocity curve increases. I thought that it was a divergence problem, but could it be due to something else? I'm sorry, but I'm not exactly sure what you mean. 

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