Timeline for in Finite Element, which approximation requires less number of unknowns: B-splines vs Shape functions vs Spectral Elements
Current License: CC BY-SA 4.0
16 events
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| Aug 24, 2023 at 21:27 | comment | added | Hosein Javanmardi | @whpowell96 You mentioned a key matter of discussion. Choosing between solving differential or integral equations or even a hybrid one is a difficult decision. According to references, I have come to this conclusion that if the matrix size is small enough, integral equation is better than PDE solving using FEM. However, as you mentioned the latter leads to sparse matrix and may be even more computationally efficient. Right now I decide to use integral equation (BEM if possible), but I am not sure if it is better than PDE/FEM. | |
| Aug 24, 2023 at 15:19 | comment | added | whpowell96 | For a steady-state linear problem that discretizes to a system $Ax=b$, solving a $1{,}000\times1{,}000$ dense system will take far more time than even a $10{,}000\times10{,}000$ system that is very sparse. This is because the typical algorithms for solving these systems in PDE applications rely only on computing the matrix-vector products $Av$ instead of constructing $A^{-1}$ directly, so the biggest bottleneck is how fast you can compute $Av$, which is faster if $A$ is sparse. | |
| Aug 24, 2023 at 15:07 | comment | added | Mikael Öhman | @HoseinJavanmardi I don't understand what you are trying to claim here. Density of a tangent matrix depends only on the level orthogonality of the chosen of basis functions. | |
| Aug 24, 2023 at 10:40 | answer | added | lightxbulb | timeline score: 3 | |
| Aug 24, 2023 at 10:04 | comment | added | Hosein Javanmardi | @MikaelÖhman If it is integral equation, the matrix is inevitably fully dense. | |
| Aug 23, 2023 at 22:27 | comment | added | Mikael Öhman | @HoseinJavanmardi Choosing basis functions that aren't (as) almost-orthogonal (e.g. for higher order continuity) would make a tangent matrix more dense, which increases the computational effort to solve the system even if the the number of unknowns are unchanged. This is just to say that your definition of "better/superior" by just counting DOFs isn't typically what anyone else would care about, but rather how long it takes to solve. | |
| Aug 23, 2023 at 22:04 | comment | added | Hosein Javanmardi | @WolfgangBangerth other metric of computational complexity is the matrix sparsity? | |
| Aug 23, 2023 at 21:11 | comment | added | lightxbulb | @HoseinJavanmardi I don't know what you mean by "better approximation profile". However if I were to take linear elements, on the same triangular mesh, and in one case I require a continuous solution, and in the other I allow discontinuities (i.e. vertices are not shared), then clearly the discontinuous space is richer, but it also has more degrees of freedom. Generally more degrees of freedom means more expensive matrix-vector multiplies due to a larger matrix bandwidth, but it also typically means faster asymptotic convergence to the solution with decreasing element size. | |
| Aug 23, 2023 at 21:04 | comment | added | lightxbulb | @ConvexHull I am specifically refering to constant elements there, I mention FVM and DG because at least I don't know of a way to make those work with continuous Galerkin. As far as higher order DG elements go, the Lagrange ones scale the same, the difference is that you get globally less dofs in CG since some nodes are shared (you can compute an upper bound e.g. for a triangulation). | |
| Aug 23, 2023 at 19:38 | comment | added | Wolfgang Bangerth | Just counting DoFs is not a useful metric. If that's your goal, use a global Fourier basis. What matters more is how long it actually takes you to compute something with the basis you choose. | |
| Aug 23, 2023 at 15:42 | comment | added | ConvexHull | @lightxbulb Could you elaborate about your claim: "one DOF per element using discontinuous Galerkin method". This is definitively not the case with higher order polynomials. | |
| Aug 23, 2023 at 13:57 | comment | added | Hosein Javanmardi | @lightxbulb I "feel" that basis functions of degree $N$ with discontinuous derivatives have better approximation profile compared to those with all its derivatives continuous. Is that correct? | |
| Aug 23, 2023 at 13:26 | comment | added | lightxbulb | The two are related. | |
| Aug 23, 2023 at 12:01 | comment | added | Hosein Javanmardi | @lightxbulb I mean, the total number of unknowns in the whole domain, not the number of unknowns per element. | |
| Aug 23, 2023 at 10:36 | comment | added | lightxbulb | The least number of unknowns is for constant elements (one per element), which is essentially finite volumes or discontinuous galerkin. However the asymptotic convergence of those is not very good. Higher order polynomials have more degrees of freedom (e.g. in 2D: 1, 3, 6, 10, 15) and thus result in more unknowns, but at the same time they offer better convergence as the size of the elements decreases. So there is no "best", it depends on your setting. | |
| Aug 23, 2023 at 8:41 | history | asked | Hosein Javanmardi | CC BY-SA 4.0 |