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January 21, 2011, 17:58 
Create a biharmonic operator

#1 
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Pascal
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Location: Montreal
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Hi all,
I would like to use a biharmonic operator for implicit calculation. I tried to use the laplacian of the laplacian but it didn't work. So should I implement a new operator? If yes, does anyone could help me to get started. I took a look at the laplacian scheme : /OpenFOAM/OpenFOAM1.6.x/src/finiteVolume/finiteVolume/laplacianSchemes/laplacianScheme and I'm a little confused with the oriented object programming. Any help would be greatly appreciated. Pascal 

August 1, 2011, 17:55 

#2 
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Mieszko Młody
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Location: POLAND, USA
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Hi Pascal,
I am facing the same problem right now. Did you were able to implement biharmonic operator ? Or maybe you know some other solution for this problem ? Thanks ZM 

August 1, 2011, 21:45 

#3 
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Pascal
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Hi,
No solution have been found yet, but if you find one let me know. Pascal 

August 2, 2011, 16:35 

#4 
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Mieszko Młody
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Hi Pascal,
The only solution I can think about is just explicit discretization. If you want to calculate laplacian(laplacian(f(x,y))) = fxxxx + 2fxxyy + fyyyy then you can make it like that: fvc::laplacian(fvc::laplacian(f)) ZM 

August 2, 2011, 17:23 

#5 
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Pascal
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Location: Montreal
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Hi Ziemowit,
I tried this one but the stability condition on such explicit discretization seems very restrictive. I think it has for consequence that you need to reduce dt a lot. No? Pascal 

August 2, 2011, 17:48 

#6 
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Mieszko Młody
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yes,
it is very possible... 

October 19, 2011, 06:28 

#7 
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Eysteinn Helgason
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Hi all,
Have you managed to get working the laplacian(laplacian)? Using the explicit formulation my simple case blows up very quickly. /Eysteinn 

October 24, 2011, 10:29 

#8 
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Mieszko Młody
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Hi Eysteinn,
I think that the best solution is just to solve two equations instead on one: laplacian(laplacian(f)) = g can be solves as: 1) laplacian(h) = g 2) laplacian(f) = h ZM 

October 31, 2011, 08:50 

#9  
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Eysteinn Helgason
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Quote:
Thank you for the answer and sorry for my late answer. This does not seem to solve my problem. Yes it runs but diverges for the simplest cases. btw. my equation also includes time derivative + extra terms: /Eysteinn 

October 31, 2011, 11:29 

#10 
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Mieszko Młody
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Hi ,
As far as I know OF has only linear solvers for algebraic equations. So it means that it is impossible to discretized nonlinear terms in the implicit way (using fvm:: ). In my opinion (but I am not an OF expert) it seems that in your equation: you have to treat first and third term explicitly and only second can be treated implicitly. Of course third term is notlinear but as far as I know there is no biharmonic operator in OF. In general I was dealing with biharmonic operator as well, and I solved it the way I posted before. But I was solving it in Matlab using FFT not in OF. The only idea I have now is : and because diffusionlike terms are treated explicitly then you have to pay attention to the time step, it has to be small enough to fulfill the condition: I think that it should work. I would recommend for the beginning to solve: 

October 31, 2011, 11:31 

#11 
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Mieszko Młody
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sorry, it should be:
... Of course third term is not nolinear but as far as I know there ... ... 

May 20, 2016, 19:43 

#12 
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Kareem Abdelshafy
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Location: Boston MA USA
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I think Mieszko algorithm for solving the fourth order equation is not right.
fvm::laplacian (h) =c fvm::ddt(c) = fvc::laplacian (D, h) This is equal to fvm::ddt(c) = D*c 

May 22, 2016, 06:02 

#13 
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Mieszko Młody
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yes,
you are right, it should be: h = fvc::laplacian (c) fvm::ddt(c) = fvm::laplacian (D, h) This is equal to fvm::ddt(c) = fvm::laplacian (D, fvc::laplacian (c)) 

Tags 
biharmonic, hyper viscosity, operator, scheme 
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