Menu

Discussion Forum

New Discussion
Filtern

Note: This discussion is about an older version of the COMSOL Multiphysics® software. The information provided may be out of date.

Discussion Closed This discussion was created more than 6 months ago and has been closed. To start a new discussion with a link back to this one, click here.

Solving a 4th order PDE: how to specify boundary conditions

Posted 20.12.2012, 22:06 GMT-5 Fluid & Heat, Studies & Solvers, Structural Mechanics Version 4.3a 4 Replies

ThSG Moore
Send Private Message Flag post as spam

Please login with a confirmed email address before reporting spam

A fourth order PDE like the biharmonic equation: ∇^4 w = 0 can be placed in Comsol's standard forms by writing u = ∇^2 w, so that we have system of equations:

∇^2 v = w
∇^2 w = 0

where v = v(x,y) and w = w(x,y).

Suppose that in the original problem, I had the two boundary conditions:

w = 1 on D1
dw/dn = 0 on D1

w = unknown constant on D2
dw/dn = 1 on D2

I am not sure how I translate to the system now in (v, w). In particular, the value of w on the boundary D2 is unknown, but it is known that D2 is constant.

From my reading, it seems that I have to use the notion of Lagrange multipliers. But there is very little documentation on this in Comsol's manuals. I would appreciate if someone could help.



4 Replies Last Post 23.12.2012, 15:54 GMT-5
Pu
Send Private Message Flag post as spam

Please login with a confirmed email address before reporting spam

Posted: 1 decade ago 21.12.2012, 05:18 GMT-5
Hi,

I also have a 4th-order PDE to solve, and it's more complex than the one in your question. I've been struggling with that for some time. Hope we both can solve our problems.

Here's a few thoughts about your question:

One thing to note is that the knowledge of Dirichlet boundary condition on a boundary enables you to calculate the tangential derivative of the dependent variable. So on boundary D1, dw/dt=0 (as your Dirichlet BC has a constant value) and dw/dn=0 means dw/dx=dw/dy=0, then everything about v on D1 is zero. For D2, dw/dt=0 and dw/dn=1. From these two conditions, I guess you can find the conditions for v on D2.

What I'm not sure about is which of these conditions one should impose explicitly as some of them are obviously redundant.

Best
--
Pu, ZHANG ??
Departamento de Física Teórica de la Materia Condensada,
Universidad Autónoma de Madrid,
Madrid, Spain.
Hi, I also have a 4th-order PDE to solve, and it's more complex than the one in your question. I've been struggling with that for some time. Hope we both can solve our problems. Here's a few thoughts about your question: One thing to note is that the knowledge of Dirichlet boundary condition on a boundary enables you to calculate the tangential derivative of the dependent variable. So on boundary D1, dw/dt=0 (as your Dirichlet BC has a constant value) and dw/dn=0 means dw/dx=dw/dy=0, then everything about v on D1 is zero. For D2, dw/dt=0 and dw/dn=1. From these two conditions, I guess you can find the conditions for v on D2. What I'm not sure about is which of these conditions one should impose explicitly as some of them are obviously redundant. Best -- Pu, ZHANG ?? Departamento de Física Teórica de la Materia Condensada, Universidad Autónoma de Madrid, Madrid, Spain.
ThSG Moore
Send Private Message Flag post as spam

Please login with a confirmed email address before reporting spam

Posted: 1 decade ago 21.12.2012, 09:23 GMT-5

One thing to note is that the knowledge of Dirichlet boundary condition on a boundary enables you to calculate the tangential derivative of the dependent variable. So on boundary D1, dw/dt=0 (as your Dirichlet BC has a constant value) and dw/dn=0 means dw/dx=dw/dy=0, then everything about v on D1 is zero. For D2, dw/dt=0 and dw/dn=1. From these two conditions, I guess you can find the conditions for v on D2.


Hi,

Perhaps I'm being dense, but on D1, where w = 0 and dw/dn = 0 on the boundary, why does that necessarily imply that v = 0?

For example, take the function w(x,y) = x^3 y^2 along D1 defined by y = 0. Then both w = 0 and dw/dn = dw/dy = 0 on the boundary, but the laplacian of w (and thus v) is non-zero.

Do you know much about the Lagrangian multipliers option in Comsol? I see mu = [mu1, mu2] in the equation definition in the screenshot here: i.imgur.com/LCDpx.png, but where is this used and where is it defined? It does not appear in the Neumann/Flux condition.
[QUOTE] One thing to note is that the knowledge of Dirichlet boundary condition on a boundary enables you to calculate the tangential derivative of the dependent variable. So on boundary D1, dw/dt=0 (as your Dirichlet BC has a constant value) and dw/dn=0 means dw/dx=dw/dy=0, then everything about v on D1 is zero. For D2, dw/dt=0 and dw/dn=1. From these two conditions, I guess you can find the conditions for v on D2. [/QUOTE] Hi, Perhaps I'm being dense, but on D1, where w = 0 and dw/dn = 0 on the boundary, why does that necessarily imply that v = 0? For example, take the function w(x,y) = x^3 y^2 along D1 defined by y = 0. Then both w = 0 and dw/dn = dw/dy = 0 on the boundary, but the laplacian of w (and thus v) is non-zero. Do you know much about the Lagrangian multipliers option in Comsol? I see mu = [mu1, mu2] in the equation definition in the screenshot here: http://i.imgur.com/LCDpx.png, but where is this used and where is it defined? It does not appear in the Neumann/Flux condition.
Pu
Send Private Message Flag post as spam

Please login with a confirmed email address before reporting spam

Posted: 1 decade ago 22.12.2012, 08:11 GMT-5
Sorry, I made some mistake.

When w=const on D1, you have dw/dt=0. Together with dw/dn=0, you know dw/dx=dw/dy=0. But I think we can't get more.

Best
--
Pu, ZHANG ??
Departamento de Física Teórica de la Materia Condensada,
Universidad Autónoma de Madrid,
Madrid, Spain.
Sorry, I made some mistake. When w=const on D1, you have dw/dt=0. Together with dw/dn=0, you know dw/dx=dw/dy=0. But I think we can't get more. Best -- Pu, ZHANG ?? Departamento de Física Teórica de la Materia Condensada, Universidad Autónoma de Madrid, Madrid, Spain.
Alexander Warning
Send Private Message Flag post as spam

Please login with a confirmed email address before reporting spam

Posted: 1 decade ago 23.12.2012, 15:54 GMT-5
you don't need lagrange multipliers, just follow this example:

www.comsol.com/support/knowledgebase/816/

worked for me
you don't need lagrange multipliers, just follow this example: http://www.comsol.com/support/knowledgebase/816/ worked for me

Note that while COMSOL employees may participate in the discussion forum, COMSOL® software users who are on-subscription should submit their questions via the Support Center for a more comprehensive response from the Technical Support team.

Suggested Content