Why does COMSOL lead to zero solution for a high-order PDE?

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Pu

Pu

December 11, 2012 3:45pm UTC

Why does COMSOL lead to zero solution for a high-order PDE?

Dear all,

I'm trying to solve a system of PDE, where the highest order is 4. COMSOL supports finite elements of 5th order. So solving the equation system should be possible by using COMSOL. But the result is zero.

Boundary conditions may be a concern, as high-order PDE generally require more boundary conditions. In my case, I imposed two homogeneous Dirichlet conditions for the dependent variable (scalar) and it's divergence. These two conditions have corresponding physical meanings.

Does anybody have an idea about such a problem? I appreciate your suggestions!

Pu.

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Ivar Kjelberg

Ivar Kjelberg

December 11, 2012 7:26pm UTC in response to Pu

Re: Why does COMSOL lead to zero solution for a high-order PDE?

Hi

I'm not so sure, I believe COMSOL only supports 2nd order PDE's, but allows for up to 5th order discretization (not sure would have to check, I havent really used more than 3rd order ;)

If you have higher order (>2nd order) PDE's you need to use intermediate dependent variables, with that you migth go higher in the PDE order.

But, as you say, you need to define sufficient BC's too, slightly trickier when we come to much higher PDE orders, still possible

--
Good luck
Ivar

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Pu

Pu

December 11, 2012 7:58pm UTC in response to Ivar Kjelberg

Re: Why does COMSOL lead to zero solution for a high-order PDE?

Hi, Ivar!

Thanks for the reply!

Yes, most the PDEs in the documentation are 2nd order ones. But I don't see explicit statement about the limit of orders. I hope COMSOL can treat higher-order PDEs... I also sent this problem to technical support. Let's wait for their answer.

About the possibility of using intermediate dependent variables, I'm trying that. Now perhaps I can introduce another dependent variable and decompose the 4th-order PDE into a 1st-order and a 3rd-order PDEs. But I'm not lucky enough to get 2 2nd-order ones.

Pu.

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Ivar Kjelberg

Ivar Kjelberg

December 11, 2012 9:25pm UTC in response to Pu

Re: Why does COMSOL lead to zero solution for a high-order PDE?

Hi

I'm afraid you need to get to 2x2nd rder, there were sme threads on the Forum, perhaps 6 monts or a year ago, and there might be something in the Knowledge base try a search. But I'm rather convinced that the General PDE is limited to 2nd order level, but that normally 4th order are solved via 2 steps with intermediate dependent variables

--
Good luck
Ivar

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Pu

Pu

December 11, 2012 10:17pm UTC in response to Ivar Kjelberg

Re: Why does COMSOL lead to zero solution for a high-order PDE?

Hi, Ivar

Thanks!

I'll try to get a simple 3rd-order PDE as an example to see if it works or not.

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Pu

Pu

December 11, 2012 10:21pm UTC in response to Ivar Kjelberg

Re: Why does COMSOL lead to zero solution for a high-order PDE?

I found the thread you mentioned, but unfortunately there's not much information. Thanks anyway!

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Jeff Hiller

Jeff Hiller
COMSOL Employee
USA
Moderator

December 11, 2012 10:30pm UTC in response to Pu

Re: Why does COMSOL lead to zero solution for a high-order PDE?

This link may be useful: www.comsol.com/support/knowledgebase/816/ . It illustrates the approach that Ivar mentioned, involving intermediary variables.

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Pu

Pu

December 12, 2012 9:10am UTC in response to Jeff Hiller

Re: Why does COMSOL lead to zero solution for a high-order PDE?

Hi, Jean-Francois!

The link is indeed helpful. Now it's clear I must find a way to formulate my problem as two 2nd-order PDE's.

One more word, I thought COMSOL could handle the highest order of 4 as there are 5th-order finite elements. Obviously due to my ignorance of FEM theory, I still feel confused about why COMSOL can only handle at most 2nd-order PDE's directly. Could you please give a hint?

Thanks a lot!

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Pu

Pu

December 12, 2012 11:45am UTC in response to Jeff Hiller

Re: Why does COMSOL lead to zero solution for a high-order PDE?

Hi, Ivar and Jean-Francois!


Thanks for the suggestions from both of you!

Here | still have a related question.

In the knowledge base link, besides the method of solving 2 2nd-order PDEs another solution is also provided. See below:

Second solution (in 2D)
The weak reformulation of the PDE states that the integral of u_test*f - uxx_test*uxx - uxx_test*uyy - uyy_test*uyy equals a Lagrange multiplier term, for all test functions u_test. Thus we can solve this equation with COMSOL Multiphysics using the weak form. Since the second derivatives uxx and uyy occur in the weak expression, we need to use a discretization that allows computation of these second derivatives. That is, the variable u should be represented using finite elements of Argyris type. Select Argyris - Quintic from the Element tab in the Subdomain Settings dialog box, or enter the shape function sharg_2_5('u'). Argyris elements are only implemented in 2D, but in 1D you can use Hermite elements of order 3, shherm(1,3,'u'), instead.

So according to this second solution a PDE can be solved directly as long as the weak form only has derivatives of orders not higher than 2. Is my understanding correct?

Thanks!

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Ivar Kjelberg

Ivar Kjelberg

December 12, 2012 8:35pm UTC in response to Pu

Re: Why does COMSOL lead to zero solution for a high-order PDE?

Hi

I'm not sure we talk about the same thing then, for me a 4th order PDE requires 4 integration steps, while you are talking about a second order PDE integrated twice to get the solution, and then looking at the 2 consecutive derivatives of the dependent solution.

Your BC are then set only for the 2 first integrations, and you have none for the two derivations, so you do not cover all solutions completely, or have I misunderstood something ?

--
Good luck
Ivar

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Pu

Pu

December 12, 2012 11:16pm UTC in response to Ivar Kjelberg

Re: Why does COMSOL lead to zero solution for a high-order PDE?

Dear Ivar,

I'm sorry for causing the possible misunderstanding.

Basically my PDE is like this:
where is the dependent variable and is some known vector function. I omit all the coefficients.

In one previous message I said I could put the PDE into certain weak form. It's misleading. I didn't mean the reformulation is possible directly for my PDE. Only after introducing another dependent variable it's possible to put the two new PDEs into proper weak forms.

By the way, I would like to take this chance to consult you on the boundary conditions for the PDE. From physical point of view, both and its divergence have physical meanings and approach zero at the boundary (I choose a boundary far enough from my physical system). So the boundary condition is kind of natural, but I'm not sure which ones I should explicitly impose in COMSOL. By means of decompsing the PDE into two I think I can correctly input the PDE into COMSOL, but there's still convergence problem. The good point is that the solution (even convergence is not achieved) generally looks reasonable, except some obvious disturbance near the boundary. Perhaps my current boundary condition setting of and is not a good choice. Do you have any comment on boundary conditions?

Thanks a lot!
--
Pu, ZHANG ??
Departamento de Física Teórica de la Materia Condensada,
Universidad Autónoma de Madrid,
Madrid, Spain.

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