10 votes

Choice of basis in FEM

People use all sorts of bases in practice. For example, people use orthonormal bases in DG methods to ensure that the mass matrix in time stepping schemes is diagonal. People also use hierarchical ...
Wolfgang Bangerth's user avatar
6 votes

Finding Shape Functions for a Triangle in 3D coordinate space

This is actually quite simple. Let's say you have your (two-dimensional) reference triangle $\hat K=\left\{(\xi,\eta)\in {\mathbb R}^2: 0\le \xi \le 1, 0\le \eta\le 1, \xi \le (1-\eta)\right\}$. Then ...
Wolfgang Bangerth's user avatar
6 votes

Choice of basis in FEM

The nodal Lagrange bases are nice because they interpolate the functions at the knots: $$\phi_i(x_j) = \delta_{ij}$$ This means that you can read and plot solutions by looking only at the coefficients ...
Bill Barth's user avatar
  • 10.9k
5 votes
Accepted

Measuring the extent to which two sets of vectors span the same space

The classical tool for this job is canonical angles. The canonical angles between $\operatorname{Im} A$ and $\operatorname{Im} B$ can be computed as $\arccos \sigma_i$, where $\sigma_i$ are the ...
Federico Poloni's user avatar
4 votes

Choice of basis in FEM

In Engineering nodal bases are a good starting point for solid mechanics problems because the principle of virtual work for the discretised system $\mathbf{K} \mathbf{u} = \mathbf{f}$ reads $$ \delta \...
Stefano M's user avatar
  • 3,839
4 votes

Choice of basis in FEM

There are a variety of different bases in FEM, but most involve basis functions which are associated with topological entities, like vertices, edges, faces, and element interiors. This makes it ...
Jesse Chan's user avatar
  • 3,142
4 votes
Accepted

Definition of Lagrange nodes in Gmsh

I think you've got slightly the wrong end of the stick from the documentation. As with a lot of other software in the area, GMSH started out with low order, hard coded numberings. These are the ones ...
origimbo's user avatar
  • 2,249
3 votes

Orthonormal basis for hexahedron

On quad/hex you can use tensor product polynomials. For example in 2d, you map the cell to a reference cell $[-1,1] \times [-1,1]$ and if $N$ is the degree, you would use $$ P_i(\xi) P_j(\eta), \qquad ...
cfdlab's user avatar
  • 3,028
3 votes

FEM shape functions on triangular elements: transition from 2D to 3D

The derivatives of the basis functions with respect to the physical space are given by \begin{equation} \phi_{,x}=\phi_{,\xi}\xi_{,x}+\phi_{,\eta}\eta_{,x} \\ \phi_{,y}=\phi_{,\xi}\xi_{,y}+\phi_{,\eta}...
miwa's user avatar
  • 31
3 votes

FEM shape functions on triangular elements: transition from 2D to 3D

For the case of faceted triangle geometry, the derivatives you're looking for ($\frac{d\phi}{dx}$, $\frac{d\phi}{dy}$, $\frac{d\phi}{dz}$) can actually be found without resorting to calculus (chain ...
rchilton1980's user avatar
  • 4,862
2 votes
Accepted

Quick way to find a common basis of eigenvectors between 2 matrices : valid or not?

Yes, the eigenvectors found with this method may depend on $x$ and $y$, but no, it doesn't matter in practice. If $A$ and $B$ share a basis of common eigenvectors, then $$ A = V\operatorname{diag}(\...
Federico Poloni's user avatar
1 vote

How to perform FFT from plane-wave basis function coefficients to real space?

The whole point of the FFT is to avoid the costly matrix-vector multiplication of the DFT, which has an effort of $\mathcal O(N^2)$ (where $N = N_x N_y N_z$), and perform the unitary transformation ...
davidhigh's user avatar
  • 3,127
1 vote

Finding Shape Functions for a Triangle in 3D coordinate space

From @Wolfgang Bangerth's answer, there exists a mapping function $(x,y,z)=\Phi(\xi,\eta)$ which can be expressed as a function of the basis shape functions. Using a similar notation as @Wolfgang ...
mfnx's user avatar
  • 172
1 vote

Computing size of N-Dimensional Polynomial Basis and Efficient Representation of Basis

As for how to create a linear index for the polynomial terms, let's consider an arrangement of terms that works nice for deduction. The terms for each dimension are enumerated as $a,b,c,\dots$. A one ...
syockit's user avatar
  • 161
1 vote

Choice of basis in FEM

Ideally, one would like to have an orthonormal basis with respect to the $L_2$ or energy inner product, or both (e.g., the eigenmodes of the operator at hand), because many matrices would be diagonal ...
Oskar Limka's user avatar
1 vote

Using SVD to biorthogonalize left and right eigenvectors?

I guess, the final result is suffering from instability (relative) of Gram-Schmidt, even in modified/stabilized forms. In this case, instead of going from the start: modifying the orthogonalization ...
Anton Menshov's user avatar
  • 8,672

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