Discussion Forum

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 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

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Good luck
Ivar

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