# Tag Info

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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 bases when doing $p$ adaptivity because it makes the construction of constraints at faces where different polynomial degrees come together trivial. For higher ...

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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 $u_j$ in the representation:$$u_h(x) = \sum_j u_j\phi_j(x)$$ That's much nicer than having to evaluate the sum at every point you care about knowing $u_h$.

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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 you know how to define the three basis functions on this triangle. They are: $$\hat N_1(\xi,\eta) = 1-\xi-\eta,\\ \hat N_2(\xi,\eta) = \xi,\\ \hat N_3(\xi,\eta)... 4 There are a variety of different bases in FEM, but most involve basis functions which are associated with topological entities, like vertices, edges, faces, and element interiors. This makes it possible to enforce various types of continuity by ensuring that degrees of freedom for such functions match at shared vertices/edges/faces. These basis functions ... 4 In Engineering nodal bases are a good starting point for solid mechanics problems because the principle of virtual work for the discretised system \mathbf{K} \mathbf{u} = \mathbf{f} reads$$ \delta \mathbf{u}^T \mathbf{K} \mathbf{u} = \delta \mathbf{u}^T \mathbf{f} $$showing that \mathbf{K} \mathbf{u} are indeed (equivalent) nodal internal forces, and \... 4 I think you've got slightly the wrong end of the stick from the documentation. As with a lot of other software in the area, GMSH started out with low order, hard coded numberings. These are the ones with the ASCII art representations, which only give first and second order numberings for tetrahedra (hence there aren't any face nodes in the 4 node or 10 node "... 3 Yes, the eigenvectors found with this method may depend on x and y, but no, it doesn't matter in practice. If A and B share a basis of common eigenvectors, then$$ A = V\operatorname{diag}(\alpha_1,\dots,\alpha_n)V^{-1}, \quad B = V\operatorname{diag}(\beta_1,\dots,\beta_n)V^{-1}. $$If A has all distinct eigenvalues, then V is unique, up to ... 3 The derivatives of the basis functions with respect to the physical space are given by \begin{equation} \phi_{,x}=\phi_{,\xi}\xi_{,x}+\phi_{,\eta}\eta_{,x} \\ \phi_{,y}=\phi_{,\xi}\xi_{,y}+\phi_{,\eta}\eta_{,y} \\ \phi_{,z}=\phi_{,\xi}\xi_{,z}+\phi_{,\eta}\eta_{,z} \end{equation} Hence, you need \xi_{,x}, \xi_{,y}, \xi_{,z}, \eta_{,x}, \eta_{,y}, \... 3 For the case of faceted triangle geometry, the derivatives you're looking for (\frac{d\phi}{dx}, \frac{d\phi}{dy}, \frac{d\phi}{dz}) can actually be found without resorting to calculus (chain rule / jacobian), you can deduce them from purely geometrical considerations. These derivatives are the cartesian (x,y,z) components of the vector function \... 3 On quad/hex you can use tensor product polynomials. For example in 2d, you map the cell to a reference cell [-1,1] \times [-1,1] and if N is the degree, you would use$$ P_i(\xi) P_j(\eta), \qquad 0 \le i,j \le N $$as the basis functions, where \xi,\eta are coordinates on the reference cell and P_i is the Legendre polynomial of degree i. These are ... 1 From @Wolfgang Bangerth's answer, there exists a mapping function (x,y,z)=\Phi(\xi,\eta) which can be expressed as a function of the basis shape functions. Using a similar notation as @Wolfgang Bangerth, the reference or base triangle is defined as:$$ \left\{(\xi,\eta)\in {\mathbb R}^2: 0\le \xi \le 1, 0\le \eta\le 1, \xi \le (1-\eta)\right\} $$The ... 1 As for how to create a linear index for the polynomial terms, let's consider an arrangement of terms that works nice for deduction. The terms for each dimension are enumerated as a,b,c,\dots. A one dimensional polynomial is straightforward$$ \begin{matrix} 0 & 1 & 2 & 3 & 4 & \dots \\ \hline 1 & a & a^2 & a^3 & a^4 &...

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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 and thus very easy to invert. But that's hard to get (solving eigenproblems is a nonlinear procedure). Lagrange bases is provides sparse matrices, which in ...

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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 process, we can start with the final result instead. So, after solving a non-Hermitian eigenvalue problem (with a reliable algorithm), we obtained $X_L$ and $X_R$...

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Density functional theory optimizes the total energy with respect to the (possibly constrained) electronic density. The usual numerical procedure for this optimization is embodied by the KS method. The KS procedure will give you the optimal total energy and the ground-state (gs) density for which this optimal energy is achieved. This optimization procedure ...

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Also take a look at John Burkhardt's libraries at http://people.sc.fsu.edu/~jburkardt/ His code is LGPL licensed, as opposed to the GPL used by the Gnu Scientific Library.

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Here are some references that might be useful: http://pubs.acs.org/doi/abs/10.1021/cr00074a002 http://www.sciencedirect.com/science/article/pii/0167797785900036 http://books.google.com/books?id=Gt4pnp-UFhUC&lpg=PA853&vq=helgaker&pg=PA725#v=snippet&q=helgaker&f=false (if you see nothing, click next then prev). See http://www....

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