Timeline for Solving a linear equation system with pure Neumann condition
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| May 15, 2021 at 23:21 | comment | added | lightxbulb | @DrHansGruber The link is dead. Could you tell me the title of the paper? Nvm, used the wayback machine, it's: ON THE FINITE ELEMENT SOLUTION OF THE PURE NEUMANN PROBLEM by Pavel Bochev. | |
| Jan 21, 2017 at 23:54 | comment | added | DrHansGruber | I case someone still stumbles upon this answer, it is incorrect and Conjugate Gradients works for pure Neumann problems, see e.g. cs.sandia.gov/~pbboche/papers_pdf/2005SIREV.pdf | |
| Nov 28, 2015 at 21:50 | history | edited | Christian Clason | CC BY-SA 3.0 |
damn you, autocorrect!
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| Jul 31, 2015 at 23:06 | comment | added | James | @shuhalo If you specify one node as a dirichlet condition then I think the matrix will then be non-singular. e.g. if A =[1,-1;-1,1] and you make the first node dirichlet then now A = [1,0;-1,1] which is now non-singular. | |
| Jun 3, 2015 at 13:48 | comment | added | Kozuki | @shuhalo but the new system converged fast, does that happen to a singular matrix? | |
| Jun 3, 2015 at 7:44 | comment | added | shuhalo | The resulting matrix will still be singular, even with the additional condition added. | |
| Jun 2, 2015 at 7:20 | comment | added | Hsien-Ming Ku | Well, if the matrix is singular, maybe I suggest you use some regularization methods, such as Tikhonov regularization. | |
| Jun 2, 2015 at 4:59 | comment | added | Kozuki | Yes, it is singular, and that is why I have to add one Dirichlet bc. Because I only care about the gradient field. Maybe I can use QR solvers, but they are too slow. | |
| Jun 1, 2015 at 18:13 | history | answered | Hsien-Ming Ku | CC BY-SA 3.0 |