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biharmonic PDE

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Hi,

I am trying to solve the 4th order biharmonic PDE with COMSOL. I'm new to COMSOL and I have been able to draw the necessary 2D geometry for the set up. However, I am having trouble inputting setting up the necessary equations.

I've been following this tutorial: www.cct.lsu.edu/~neilan/Neila...es/Argyris.pdf

However, when I try the step on slide 9 of the tutorial to input the equation, the field just resets to the old value. I've been opening up other templates as well (such as the general coefficient in PDE modes) which are suppose to be for custom equations. However, it seems that these custom eqns are limited to second order PDEs, and I need to solve a 4th order PDE.

Any help on this would be much appreciated!

3 Replies Last Post 2010年12月28日 GMT-5 01:12

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Posted: 1 decade ago 2010年11月24日 GMT-5 18:06
HI
First your link is not working
second it is easy to resolve a 4 order equation within comsol use a system of two second order equation [ this is a classical trick in ODE resolution ]
so for solving say uxxxx=f you write uxx=v and vxx=f and comsol can solve this for (u,v) [ with the bc that make it a well posed problem of course.].

good luck

jf
HI First your link is not working second it is easy to resolve a 4 order equation within comsol use a system of two second order equation [ this is a classical trick in ODE resolution ] so for solving say uxxxx=f you write uxx=v and vxx=f and comsol can solve this for (u,v) [ with the bc that make it a well posed problem of course.]. good luck jf

Ivar KJELBERG COMSOL Multiphysics(r) fan, retired, former "Senior Expert" at CSEM SA (CH)

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Posted: 1 decade ago 2010年11月25日 GMT-5 01:19
Hi Jean François

Well back Iassume ;)

I would add to your comments, COMSOL is set up only for 2nd order BVP, but as you explain any higher order can be reached via additional variables, so long we stay within well behaved BVP with sufficient boundary definitions to have a solvable system.

There are finally very few (if any) common physical problems that cannot be solved with the equation set of COMSOL, Allthough one can imagine many mathematical equation sets not solvable like that, with COMSOL

--
Good luck
Ivar
Hi Jean François Well back Iassume ;) I would add to your comments, COMSOL is set up only for 2nd order BVP, but as you explain any higher order can be reached via additional variables, so long we stay within well behaved BVP with sufficient boundary definitions to have a solvable system. There are finally very few (if any) common physical problems that cannot be solved with the equation set of COMSOL, Allthough one can imagine many mathematical equation sets not solvable like that, with COMSOL -- Good luck Ivar

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Posted: 1 decade ago 2010年12月28日 GMT-5 01:12
Hi guys,

Thanks for the help. I have been playing around with COMSOL and am now a bit more familiar how to do coupled equations. I am still having trouble getting the right solution for a PDE with known solution, so I'd like to specifically lay out the problem here.

SHORT: How do I implement Cauchy bcs in COMSOL without this lagrange multiplier term in the Dirichlet bc form? Also, to solve just the biharmonic eqn in a circular disc with a hole in the middle, is there any need to fiddle with the parameters in the weak formulation options (of subdomain settings and boundary condition settings)?

LONG: The problem at hand is grad^4(u) = 0, so we decouple and as you guys have said, obtain grad^2(u) = k & grad^2(k) = 0. The domain is the entire real plane with a circular hole of radius 1 centered at (0,0). To approximate the entire real domain, I am just using a bigger circle of radius 100, so the domain is really just a big circle with a small hole at the center.

The boundary conditions at r = 1 are u(r=1) = 0, n*grad(u)(r=1) = -x, where x is a constant. The boundary conditions at r->Inf (i.e. r = 100) is n*grad(u)(r=100) = 0. If one solves this problem analytically, the only radially symmetric eigenfunction that satisfies the BCs is -x * log(r). However, with COMSOL, I am just getting 0, so something seems wrong.

For the exterior circle (r= 100) I use the Neumann bcs as I just need to specify the normal derivative. For the interior (r=100), I have Cauchy bcs, and need to specify both u and n*grad(u). So far it seems that I need to use the Dirichlet bcs, which has the term R = [0 0], so I set R = [u 0] to take care of the u(r=1) = 0 bc. However, then there is a -n*grad[-u -k] = G + (dR/du) ^ T * mu term, where mu is a lagrange multiplier. I set G = [0, kx+ky] since -n*grad[-u -k] = [0, kx+ky] at r = 100. However, this last term with the lagrange multiplier troubles me. According to help documentation this lagrange multiplier adjusts so all bcs are satisfied, but I know the initial bvp is well posed, so I feel that there should be no extra term lagrange multiplier term .

So my question is, is there any way to get around having this lagrange multiplier term in order to implement Cauchy bcs. Also, throughout this time, I have been ignoring the weak term tab in the subdomain settings and boundary condition settings. From my understanding this is an additional part of finite element analysis, but do I need to fiddle with this to solve this basic PDE?

Thanks,
John Smith
Hi guys, Thanks for the help. I have been playing around with COMSOL and am now a bit more familiar how to do coupled equations. I am still having trouble getting the right solution for a PDE with known solution, so I'd like to specifically lay out the problem here. SHORT: How do I implement Cauchy bcs in COMSOL without this lagrange multiplier term in the Dirichlet bc form? Also, to solve just the biharmonic eqn in a circular disc with a hole in the middle, is there any need to fiddle with the parameters in the weak formulation options (of subdomain settings and boundary condition settings)? LONG: The problem at hand is grad^4(u) = 0, so we decouple and as you guys have said, obtain grad^2(u) = k & grad^2(k) = 0. The domain is the entire real plane with a circular hole of radius 1 centered at (0,0). To approximate the entire real domain, I am just using a bigger circle of radius 100, so the domain is really just a big circle with a small hole at the center. The boundary conditions at r = 1 are u(r=1) = 0, n*grad(u)(r=1) = -x, where x is a constant. The boundary conditions at r->Inf (i.e. r = 100) is n*grad(u)(r=100) = 0. If one solves this problem analytically, the only radially symmetric eigenfunction that satisfies the BCs is -x * log(r). However, with COMSOL, I am just getting 0, so something seems wrong. For the exterior circle (r= 100) I use the Neumann bcs as I just need to specify the normal derivative. For the interior (r=100), I have Cauchy bcs, and need to specify both u and n*grad(u). So far it seems that I need to use the Dirichlet bcs, which has the term R = [0 0], so I set R = [u 0] to take care of the u(r=1) = 0 bc. However, then there is a -n*grad[-u -k] = G + (dR/du) ^ T * mu term, where mu is a lagrange multiplier. I set G = [0, kx+ky] since -n*grad[-u -k] = [0, kx+ky] at r = 100. However, this last term with the lagrange multiplier troubles me. According to help documentation this lagrange multiplier adjusts so all bcs are satisfied, but I know the initial bvp is well posed, so I feel that there should be no extra term lagrange multiplier term . So my question is, is there any way to get around having this lagrange multiplier term in order to implement Cauchy bcs. Also, throughout this time, I have been ignoring the weak term tab in the subdomain settings and boundary condition settings. From my understanding this is an additional part of finite element analysis, but do I need to fiddle with this to solve this basic PDE? Thanks, John Smith

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