Timeline for Poisson equation, stiffness matrix positive definiteness, Dirichlet boundary conditions
Current License: CC BY-SA 4.0
13 events
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| Feb 17, 2021 at 21:36 | comment | added | lightxbulb | @AbdullahAliSivas Thank you. Feel free to formulate this as an answer if you want so that I can accept it. | |
| Feb 17, 2021 at 21:30 | comment | added | Abdullah Ali Sivas | Yes. $a(u,v)$ is not coercive on $H^1$, but it is coercive on $H^1_0$ (or in general, $H^1_g$ if $u=g$ on $\Gamma_D$) . | |
| Feb 17, 2021 at 15:34 | comment | added | lightxbulb | @AbdullahAliSivas I understand that the solution is defined up to a constant if no extra constraint is imposed. I am fine with that. I was more interested in the details. From what I have read, the bilinear form $a$ due to the Poisson equation applied to functions from $\mathcal{H}^1_{\Gamma_D}(\Omega)$ is in fact coercive. As I understand it, the functions $c$ simply isn't from $\mathcal{H}^1_{\Gamma_D}(\Omega)$ since it doesn't vanish on the Dirichlet boundary. That's my understanding as to why $a$ here is not coercive. Is that correct? | |
| Feb 17, 2021 at 15:24 | comment | added | Abdullah Ali Sivas | Not really. Consider the Poisson problem with only Neumann boundary conditions. Let $u$ solve that equation and $c\neq 0$ be a constant function. Then $u+c$ also solves that equation. Consider $a(c,c) = a(u+c-u,u+c-u) = a(u+c,u+c) - a(u,u) = 0$, which immediately shows us that $a(\cdot,\cdot)$ is not coercive, hence, the corresponding coefficient matrix is not positive definite (though it is positive semidefinite). By adding the extra condition $\int u =0$, you guarantee that $u+c$ is not solution, hence, $a(u,v)$ becomes coercive. Does that make sense? | |
| Feb 17, 2021 at 12:21 | comment | added | lightxbulb | @AbdullahAliSivas Is the point of your example to show a function $c$ which is not in $\mathcal{H}^1_{\Gamma_D}(\Omega)$ since it doesn't vanish on a Dirichlet boundary, and is instead only in $\mathcal{H}^1(\Omega)$. Then due to $V \subset \mathcal{H}^1$ it follows $\pmb{1}^T\pmb{W}\pmb{1} = \pmb{0}$ (if $c = 1$). I am still struggling to understand how this affects $\pmb{W}|_{Ind \times Ind}$. Is it because by introducing Dirichlet nodes we modify the functions $\phi_i$ by requiring those to vanish at node not in $Ind$, so we get $\pmb{W}|_{Ind \times Ind}$ to be positive definite from that? | |
| Feb 17, 2021 at 1:17 | comment | added | Abdullah Ali Sivas | Also in that answer, they choose their spaces carefully, namely $u,v\in H_{0}^1(\Omega)$ so Dirichlet boundary conditions are enforced implicitly -which is the standard procedure for most FE methods. | |
| Feb 17, 2021 at 1:15 | comment | added | Abdullah Ali Sivas | I guess I should have written the sentence "the nullspace only contains the unit element" differently. Using the notation in the accepted answer you linked, if there are no Dirichlet conditions, it is easy to show that $a(c,c)=0$ where $c$ is a non-zero constant function. Therefore, $a(u,v)$ is not coercive. You can similarly show that $\mathbf{1}^TW\mathbf{1}=0$, hence, $W$ is not positive definite in the absence of Dirichlet boundary conditions. | |
| Feb 17, 2021 at 0:29 | history | edited | lightxbulb | CC BY-SA 4.0 |
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| Feb 17, 2021 at 0:24 | history | edited | lightxbulb | CC BY-SA 4.0 |
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| Feb 17, 2021 at 0:18 | history | edited | lightxbulb | CC BY-SA 4.0 |
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| Feb 17, 2021 at 0:10 | comment | added | lightxbulb | @AbdullahAliSivas I am talking specifically about the matrix $\pmb{W}|_{Ind \times Ind}$. Is there a detailed formal proof of how introducing a Dirichlet node modifies this matrix in order to make it positive definite? I am missing that specific part. I also could not find at which point this detail is touched upon in: scicomp.stackexchange.com/questions/21423/… | |
| Feb 16, 2021 at 23:37 | comment | added | Abdullah Ali Sivas | Short answer: You are right, in the absence of Dirichlet b.c.s Poisson problem is only positive semi-definite, and you have to add another condition to guarantee uniqueness of the solution. However, notice that, even if you don't add any other conditions, the solution is unique up to an additive constant and the nullspace only contains the unit element. | |
| Feb 16, 2021 at 22:45 | history | asked | lightxbulb | CC BY-SA 4.0 |