# Tag Info

Accepted

### What is the global problem in the two-level additive Schwarz?

Continuous finite elements Typically, if $A$ is your finite element discretization on the finest mesh, $A_i = R_i * A * R_i^T$. So, for $i=0$, $A_0$ corresponds to the finite element discretization ...
• 607

### "Optimal" domain partitioning in domain decomposition algorithms

We use domain decomposition because we want to exploit the power of more than one processor. As a consequence, the right question to pose is: "How do we need to partition the domain so that we ...
• 56.2k
Accepted

### What is the difference between Adittive Schwarz as a preprocessor and a solver?

By itself, Schwarz methods are stationary iterations just like Jacobi, Gauss-Seidel, or SOR. They converge to the solution, but often quite slowly. But, like any other stationary method, one iteration ...
• 56.2k
Accepted

### Partition mesh into predetermined submeshes

For your first question, constructing the adjacency graph of the "partitions" (what you call "cell groups"): Let's say you have an array $p_K$ in which you store for each cell $K$ ...
• 56.2k

### Solve wave equation with discontinuous coefficients numerically?

$c$ depending on time is not the issue. You will use an RK scheme which takes care of this. The issue is $c$ is discontinuous in $x$. I recommend SBP-SAT schemes for this. (1) Derive an energy ...
• 3,038
Accepted

### Direct solvers and domain decomposition for FEM

As I understand, this is a fairly popular approach. Direct solvers are usually more efficient than iterative solvers for < 100,000 unknowns, so you partition the problem into subproblems of roughly ...
• 10.4k

### Efficient evaluation of $BQ^{-1}B^T$ (Domain Decomposition Implementation)

This is always going to be a somewhat costly calculation because the inverse of any "interesting" sparse matrix is generally dense and therefore so is $\mathbf M$. That said, there are smarter ...
• 5,046

### How to precondition FEM problems using domain decomposition?

Don't use domain decomposition methods. They're from the 1990s, but we have much better ways of preconditioning problems today. All of them work on the global problem, rather than ones on subdomains. ...
• 56.2k

### How to precondition FEM problems using domain decomposition?

Let us consider the abstract linear problem $$\mathcal{A} x = b \,,$$ where $\mathcal{A}$ is a linear operator and $x$ and $b$ some functions on a certain domain. To answer your question let me ...
• 558
Accepted

### Solve wave equation with discontinuous coefficients numerically?

Here is a brute force solution that would work no matter what is the discontinuity and nonlinearity in $c(x,t)$. Write your PDE as a system of two: $\dot{y}=z\\ \dot{z}=c^2(x,t) y_{xx}$ Now, ...
• 2,595

### preconditioner for $u''(x)=\sin(x)$

In general, the appropriate preconditioners for elliptic problems such as yours are multigrid methods. In this 1d case, however, the simplest discretizations lead to tri-diagonal matrices and in that ...
• 56.2k
1 vote

### can you give me some information of tools for load reblance

I found the ParMetis have what I want and easy to use.
• 323
1 vote

### Parallelisation strategies for mixed FE formulations

It's a misunderstanding that you need two different meshes: The proper way to see things is that you are using the same mesh, but different polynomial spaces for the two variables. For example, for ...
• 56.2k
1 vote

### interface value on the error equation

$\delta_n \neq \lambda_n - \lambda_{n-1}$. It is actually defined in (2.8) and (I don't know why) $\delta_2 = \lambda_2 - \lambda_1$. You can confirm it by subtracting first three lines of (2.2) from (...
• 2,851
1 vote

### problem in interface operator

Here's the flow of logic, hopefully in a slightly more readable form: We write the transmission conditions with as-yet-unknown linear operators $\mathcal{S}_1$ and $\mathcal{S}_2$ which operate on ...
• 1,253
1 vote
Accepted

### Partitioning SPD matrix with METIS to preserve block SPD-ness

The block SPD-ness is preserved because any partitioning does not change the SPD property of the matrix as it would not alter the eigenvalues as partitioning would involve row and column swaps which ...
• 111

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