4

$h$ is a measure of the mesh size. In the example, they are using rectangular elements. For which a commonly used measure of the mesh size is the length of the largest diagonal. Looking at the table above the sentence you quoted: $(M_5-M_6)/(M_4-M_5)=(0.14052586 −0.14056422)/(0.14037251−0.14052586)\approx 1/4 = (h_6/h_5)^2$ where $M_i$ is the mean at $i$-th ...


4

For both schemes: I am going to lump $B$ and $C$ into a single matrix $B$ for convenience of communicating the idea. Likewise, $y$ and $z$ are now $y$, and $F_y$ and $F_z$ are $F_y$. $Ax+B^Ty=F_x$ or $x = A^{-1}(F_x-B^Ty)$ $Bx=F_y$ or $BA^{-1}(F_x-B^Ty)=F_y$ $BA^{-1}B^Ty=BA^{-1}F_x-F_y$ Note that $BA^{-1}B^T$ is symmetric and invertible, and can be ...


3

Using the typical expansion functions (1-forms/edge-elements for E, and 2-forms/facet-elements for B) the formulations are basically the same after spatial discretization and you'd expect more or less the same accuracy. I do think they express slightly different opinions about time integration. The mixed E/B formulation nudges you in the direction of ...


2

Here is at least an idea, whether it works is a different question. Let's say you sort unknowns so that you have the ones in the interior of the domain first, and then all those at the boundary. Then the matrix that corresponds to your problem decomposes in the following way: $$ A = \begin{pmatrix} A^{\circ\circ} & B^{\circ\partial} \\ C^{\...


2

The elemental stiffness matrix must be always singular because while deriving it we do not impose any constraints or boundary conditions. Thus inverting the stiffness matrix to solve for displacements/position-vectors/degrees-of-freedom should yield indeterminate results.


2

Disclaimer: I wrote this answer in a rush, may not be up to the standards. Also, I don't have enough information about your problem to give more detailed advice. However, the information presented here should be enough to get you going. I will suggest you to use MINRES with an appropriate preconditioner. For a moment, let's assume that this is not a twofold ...


2

You could find the shape functions proposing a quadratic function of the form $$N_i = a_0 + a_1 r + a_2 s + a_3 rs + a_4 r^2 + a_5 s^2\, ,$$ and enforcing the following condition $$N_i(r_j, s_j) = \delta_{ij}\, ,$$ that is, 1 for each node and 0 for the others. I am using as reference element a triangle with vertices $(0, 0), (0, 1), (1, 0)$, and assuming ...


1

If you want to use the coefficient of the form $$ \alpha = \begin{cases} \alpha_1 & r < 0.5 \\ \alpha_2 & r > 0.5 \end{cases} $$ it is useful to work in polar coordinates. You build two pieces of the solution $$ u_1(r,\theta), \qquad r < 0.5 $$ and $$ u_2(r,\theta), \qquad r > 0.5 $$ Then at the interface you ensure solution and flux ...


1

If you want to know the why and the how, you probably want to watch lecture 21.65 here: https://www.math.colostate.edu/~bangerth/videos.html Which you probably want to do after you've watched lecture 21.6. Disclaimer: I'm the one in these videos.


Only top voted, non community-wiki answers of a minimum length are eligible