Difference between revisions of "Adaptivity"
From Medusa: Coordinate Free Mehless Method implementation
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This adaptive cycle below is repeated until the convergence criterion is met. The procedure on $j$-th iteration is written in more detail below: | This adaptive cycle below is repeated until the convergence criterion is met. The procedure on $j$-th iteration is written in more detail below: | ||
| − | + | <ol> | |
| − | + | <li>Fill $\Omega$ with nodes conforming to $h^j$.</li> | |
| − | + | <li>Solve the problem to obtain $u^j$.</li> | |
| − | + | <li>Compute the error indicator values $\varepsilon_i^j$ for each node $p_i$.</li> | |
| − | + | <li>If the mean of $\varepsilon_i^j$ is below some tolerance $\varepsilon$ return $u^j$ as the solution and stop.</li> | |
| + | <li>Adapt $h^j$ to obtain $h^{j+1}$.</li> | ||
| + | </ol> | ||
More details can be found in our paper: [https://arxiv.org/abs/1811.10368 https://arxiv.org/abs/1811.10368] | More details can be found in our paper: [https://arxiv.org/abs/1811.10368 https://arxiv.org/abs/1811.10368] | ||
Go back to [[Medusa#Examples|Examples]]. | Go back to [[Medusa#Examples|Examples]]. | ||
Revision as of 13:10, 11 June 2019
Go back to Examples.
The adaptive methodology in this paper behaves similarly to "remeshing" used commonly in FEM. Some initial (possibly variable) nodal spacing $h^0$ is chosen, as well as its lower and upper bounds $h_L$ and $h_U$, respectively. 3 Domain $\Omega$ is filled with nodes, conforming to $h^0$ and the solution $u^0$ is obtained. An error indicator is employed to determine which nodes should be (de)refined and the nodal density $h^0$ is altered appropriately. This adaptive cycle below is repeated until the convergence criterion is met. The procedure on $j$-th iteration is written in more detail below:
- Fill $\Omega$ with nodes conforming to $h^j$.
- Solve the problem to obtain $u^j$.
- Compute the error indicator values $\varepsilon_i^j$ for each node $p_i$.
- If the mean of $\varepsilon_i^j$ is below some tolerance $\varepsilon$ return $u^j$ as the solution and stop.
- Adapt $h^j$ to obtain $h^{j+1}$.
More details can be found in our paper: https://arxiv.org/abs/1811.10368
Go back to Examples.
