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Create a biharmonic operator

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Old   January 21, 2011, 17:58
Default Create a biharmonic operator
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Pascal
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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/OpenFOAM-1.6.x/src/finiteVolume/finiteVolume/laplacianSchemes/laplacianScheme

and I'm a little confused with the oriented object programming. Any help would be greatly appreciated.

Pascal
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Old   August 1, 2011, 17:55
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Mieszko Młody
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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
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Old   August 1, 2011, 21:45
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Hi,

No solution have been found yet, but if you find one let me know.
Pascal
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Old   August 2, 2011, 16:35
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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
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Old   August 2, 2011, 17:23
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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
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Old   August 2, 2011, 17:48
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yes,
it is very possible...
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Old   October 19, 2011, 06:28
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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
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Old   October 24, 2011, 10:29
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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
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Old   October 31, 2011, 08:50
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Eysteinn Helgason
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Quote:
Originally Posted by ziemowitzima View Post
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

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
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Old   October 31, 2011, 11:29
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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 non-linear terms in the implicit way (using fvm:: ).
In my opinion (but I am not an OF expert) it seems that in your equation:
\frac{\partial c}{\partial t} = D\nabla^2 c^3 - D\nabla^2 c - D\gamma\nabla^4 c

you have to treat first and third term explicitly and only second can be treated implicitly. Of course third term is not-linear 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 :
fvm::\nabla^2 h = c
fvm::\frac{\partial c}{\partial t} = fvc::D\nabla^2 c^3 - fvm::D\nabla^2 c - fvc::D\gamma\nabla^2 h

and because diffusion-like terms are treated explicitly then you have to pay attention to the time step, it has to be small enough to fulfill the condition:
\frac{D\Delta t}{\Delta x^2} < C, \: \frac{\gamma D\Delta t}{\Delta x^2} < C

I think that it should work. I would recommend for the beginning to solve:
fvm::\nabla^2 h = c
fvm::\frac{\partial c}{\partial t} = - fvc::D\nabla^2 h
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Old   October 31, 2011, 11:31
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sorry, it should be:
...
Of course third term is not nolinear but as far as I know there ...
...
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