Create a biharmonic operator

Pascal_doran · January 21, 2011, 17:58

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

ziemowitzima · August 1, 2011, 18:55

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

Pascal_doran · August 1, 2011, 22:45

Hi,

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

ziemowitzima · August 2, 2011, 17:35

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

Pascal_doran · August 2, 2011, 18:23

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

ziemowitzima · August 2, 2011, 18:48

yes,
it is very possible...

eysteinn · October 19, 2011, 07:28

Hi all,

Have you managed to get working the laplacian(laplacian)?

Using the explicit formulation my simple case blows up very quickly.

/Eysteinn

ziemowitzima · October 24, 2011, 11:29

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

eysteinn · October 31, 2011, 08:50

Quote:

Originally Posted by ziemowitzima

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

ziemowitzima · October 31, 2011, 11:29

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:
$fvm::\nabla^2 h = c$
$fvm::\frac{\partial c}{\partial t} = - fvc::D\nabla^2 h$

ziemowitzima · October 31, 2011, 11:31

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

Kareem Abdelshafy · May 20, 2016, 20:43

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

ziemowitzima · May 22, 2016, 07:02

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))

January 21, 2011, 17:58	Create a biharmonic operator	#1
Pascal_doran Member Pascal Join Date: Jun 2009 Location: Montreal Posts: 65 Rep Power: 17	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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August 1, 2011, 18:55		#2
ziemowitzima Senior Member Mieszko Młody Join Date: Mar 2009 Location: POLAND, USA Posts: 145 Rep Power: 17	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, 22:45		#3
Pascal_doran Member Pascal Join Date: Jun 2009 Location: Montreal Posts: 65 Rep Power: 17	Hi, No solution have been found yet, but if you find one let me know. Pascal

August 2, 2011, 17:35		#4
ziemowitzima Senior Member Mieszko Młody Join Date: Mar 2009 Location: POLAND, USA Posts: 145 Rep Power: 17	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, 18:23		#5
Pascal_doran Member Pascal Join Date: Jun 2009 Location: Montreal Posts: 65 Rep Power: 17	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, 18:48		#6
ziemowitzima Senior Member Mieszko Młody Join Date: Mar 2009 Location: POLAND, USA Posts: 145 Rep Power: 17	yes, it is very possible...

October 19, 2011, 07:28		#7
eysteinn Member Eysteinn Helgason Join Date: Sep 2009 Location: Gothenburg, Sweden Posts: 53 Rep Power: 17	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, 11:29		#8
ziemowitzima Senior Member Mieszko Młody Join Date: Mar 2009 Location: POLAND, USA Posts: 145 Rep Power: 17	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, 11:29		#10
ziemowitzima Senior Member Mieszko Młody Join Date: Mar 2009 Location: POLAND, USA Posts: 145 Rep Power: 17	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: $fvm::\nabla^2 h = c$ $fvm::\frac{\partial c}{\partial t} = - fvc::D\nabla^2 h$

October 31, 2011, 11:31		#11
ziemowitzima Senior Member Mieszko Młody Join Date: Mar 2009 Location: POLAND, USA Posts: 145 Rep Power: 17	sorry, it should be: ... Of course third term is not nolinear but as far as I know there ... ...

May 20, 2016, 20:43		#12
Kareem Abdelshafy New Member Kareem Abdelshafy Join Date: Feb 2016 Location: Boston MA USA Posts: 4 Rep Power: 10	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, 07:02		#13
ziemowitzima Senior Member Mieszko Młody Join Date: Mar 2009 Location: POLAND, USA Posts: 145 Rep Power: 17	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))